From 3d6baa3ae3648ef41e7a96bd49cb68c3d9d2df2d Mon Sep 17 00:00:00 2001 From: dcj_hp <294487055@qq.com> Date: Fri, 20 May 2022 00:14:16 +0800 Subject: [PATCH 001/136] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=20java=20=E4=B8=80?= =?UTF-8?q?=E7=BB=B4dp=E6=95=B0=E7=BB=84=E8=A7=A3=E6=B3=95=E4=BB=A3?= =?UTF-8?q?=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1143.最长公共子序列.md | 44 +++++++++++++++++++++++++++++++++ 1 file changed, 44 insertions(+) diff --git a/problems/1143.最长公共子序列.md b/problems/1143.最长公共子序列.md index ecedf89b..b4b8e6db 100644 --- a/problems/1143.最长公共子序列.md +++ b/problems/1143.最长公共子序列.md @@ -129,6 +129,9 @@ public: Java： ```java +/* + 二维dp数组 +*/ class Solution { public int longestCommonSubsequence(String text1, String text2) { int[][] dp = new int[text1.length() + 1][text2.length() + 1]; // 先对dp数组做初始化操作 @@ -146,6 +149,47 @@ class Solution { return dp[text1.length()][text2.length()]; } } + + + +/** + 一维dp数组 +*/ +class Solution { + public int longestCommonSubsequence(String text1, String text2) { + int n1 = text1.length(); + int n2 = text2.length(); + + // 多从二维dp数组过程分析 + // 关键在于 如果记录 dp[i - 1][j - 1] + // 因为 dp[i - 1][j - 1] dp[j - 1] <=> dp[i][j - 1] + int [] dp = new int[n2 + 1]; + + for(int i = 1; i <= n1; i++){ + + // 这里pre相当于 dp[i - 1][j - 1] + int pre = dp[0]; + for(int j = 1; j <= n2; j++){ + + //用于给pre赋值 + int cur = dp[j]; + if(text1.charAt(i - 1) == text2.charAt(j - 1)){ + //这里pre相当于dp[i - 1][j - 1] 千万不能用dp[j - 1] !! + dp[j] = pre + 1; + } else{ + // dp[j] 相当于 dp[i - 1][j] + // dp[j - 1] 相当于 dp[i][j - 1] + dp[j] = Math.max(dp[j], dp[j - 1]); + } + + //更新dp[i - 1][j - 1], 为下次使用做准备 + pre = cur; + } + } + + return dp[n2]; + } +} ``` Python： From 3605046879d02c5f6b931f7440935ef4142ef121 Mon Sep 17 00:00:00 2001 From: yalexu <61576631+alexgy1@users.noreply.github.com> Date: Fri, 3 Jun 2022 18:24:51 +0800 Subject: [PATCH 002/136] Add JS correct reverse method --- problems/0344.反转字符串.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index 58bada05..192a397c 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -190,13 +190,13 @@ javaScript: * @return {void} Do not return anything, modify s in-place instead. */ var reverseString = function(s) { - return s.reverse(); + //Do not return anything, modify s in-place instead. + reverse(s) }; -var reverseString = function(s) { +var reverse = function(s) { let l = -1, r = s.length; while(++l < --r) [s[l], s[r]] = [s[r], s[l]]; - return s; }; ``` From a2a2cf60272094c3c9d962723132502d93db097a Mon Sep 17 00:00:00 2001 From: lizhendong128 Date: Fri, 3 Jun 2022 21:42:36 +0800 Subject: [PATCH 003/136] =?UTF-8?q?=E4=BF=AE=E6=94=B90106=E4=BB=8E?= =?UTF-8?q?=E4=B8=AD=E5=BA=8F=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?=E5=BA=8F=E5=88=97=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=20Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 将0106和0105的Java版本进行了修改，采用了map来存储位置信息，加快定位；并且代码更容易看懂 --- .../0106.从中序与后序遍历序列构造二叉树.md | 86 +++++++++---------- 1 file changed, 39 insertions(+), 47 deletions(-) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 188ad3cb..878c3572 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -584,35 +584,29 @@ tree2 的前序遍历是[1 2 3]， 后序遍历是[3 2 1]。 ```java class Solution { + Map map; // 方便根据数值查找位置 public TreeNode buildTree(int[] inorder, int[] postorder) { - return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length); + map = new HashMap<>(); + for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置 + map.put(inorder[i], i); + } + + return findNode(inorder, 0, inorder.length, postorder,0, postorder.length); // 前闭后开 } - public TreeNode buildTree1(int[] inorder, int inLeft, int inRight, - int[] postorder, int postLeft, int postRight) { - // 没有元素了 - if (inRight - inLeft < 1) { + + public TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) { + // 参数里的范围都是前闭后开 + if (inBegin >= inEnd || postBegin >= postEnd) { // 不满足左闭右开，说明没有元素，返回空树 return null; } - // 只有一个元素了 - if (inRight - inLeft == 1) { - return new TreeNode(inorder[inLeft]); - } - // 后序数组postorder里最后一个即为根结点 - int rootVal = postorder[postRight - 1]; - TreeNode root = new TreeNode(rootVal); - int rootIndex = 0; - // 根据根结点的值找到该值在中序数组inorder里的位置 - for (int i = inLeft; i < inRight; i++) { - if (inorder[i] == rootVal) { - rootIndex = i; - break; - } - } - // 根据rootIndex划分左右子树 - root.left = buildTree1(inorder, inLeft, rootIndex, - postorder, postLeft, postLeft + (rootIndex - inLeft)); - root.right = buildTree1(inorder, rootIndex + 1, inRight, - postorder, postLeft + (rootIndex - inLeft), postRight - 1); + int rootIndex = map.get(postorder[postEnd - 1]); // 找到后序遍历的最后一个元素在中序遍历中的位置 + TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点 + int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数，用来确定后序数列的个数 + root.left = findNode(inorder, inBegin, rootIndex, + postorder, postBegin, postBegin + lenOfLeft); + root.right = findNode(inorder, rootIndex + 1, inEnd, + postorder, postBegin + lenOfLeft, postEnd - 1); + return root; } } @@ -622,31 +616,29 @@ class Solution { ```java class Solution { + Map map; public TreeNode buildTree(int[] preorder, int[] inorder) { - return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1); - } - - public TreeNode helper(int[] preorder, int preLeft, int preRight, - int[] inorder, int inLeft, int inRight) { - // 递归终止条件 - if (inLeft > inRight || preLeft > preRight) return null; - - // val 为前序遍历第一个的值，也即是根节点的值 - // idx 为根据根节点的值来找中序遍历的下标 - int idx = inLeft, val = preorder[preLeft]; - TreeNode root = new TreeNode(val); - for (int i = inLeft; i <= inRight; i++) { - if (inorder[i] == val) { - idx = i; - break; - } + map = new HashMap<>(); + for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置 + map.put(inorder[i], i); } - // 根据 idx 来递归找左右子树 - root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft), - inorder, inLeft, idx - 1); - root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight, - inorder, idx + 1, inRight); + return findNode(preorder, 0, preorder.length, inorder, 0, inorder.length); // 前闭后开 + } + + public TreeNode findNode(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd) { + // 参数里的范围都是前闭后开 + if (preBegin >= preEnd || inBegin >= inEnd) { // 不满足左闭右开，说明没有元素，返回空树 + return null; + } + int rootIndex = map.get(preorder[preBegin]); // 找到前序遍历的第一个元素在中序遍历中的位置 + TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点 + int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数，用来确定前序数列的个数 + root.left = findNode(preorder, preBegin + 1, preBegin + lenOfLeft + 1, + inorder, inBegin, rootIndex); + root.right = findNode(preorder, preBegin + lenOfLeft + 1, preEnd, + inorder, rootIndex + 1, inEnd); + return root; } } From af1b3e3e32676d6037645c0553b4fc2bb2cf149d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E6=A1=9C=E5=B0=8F=E8=B7=AF=E4=B8=83=E8=91=89?= <20304773@qq.com> Date: Sat, 4 Jun 2022 00:22:14 +0800 Subject: [PATCH 004/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880209.?= =?UTF-8?q?=E9=95=BF=E5=BA=A6=E6=9C=80=E5=B0=8F=E7=9A=84=E5=AD=90=E6=95=B0?= =?UTF-8?q?=E7=BB=84.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0=20C#=20?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0209.长度最小的子数组.md | 23 ++++++++++++++++++++++- 1 file changed, 22 insertions(+), 1 deletion(-) diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index fbef7692..ceb06ce9 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -448,6 +448,27 @@ object Solution { } } ``` - +C#: +```csharp +public class Solution { + public int MinSubArrayLen(int s, int[] nums) { + int n = nums.Length; + int ans = int.MaxValue; + int start = 0, end = 0; + int sum = 0; + while (end < n) { + sum += nums[end]; + while (sum >= s) + { + ans = Math.Min(ans, end - start + 1); + sum -= nums[start]; + start++; + } + end++; + } + return ans == int.MaxValue ? 0 : ans; + } +} +``` -----------------------
From cccb4973aeccde06a00e5c693510c824ae98c99b Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 4 Jun 2022 10:44:25 +0800 Subject: [PATCH 005/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880035.?= =?UTF-8?q?=E6=90=9C=E7=B4=A2=E6=8F=92=E5=85=A5=E4=BD=8D=E7=BD=AE.md?= =?UTF-8?q?=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0035.搜索插入位置.md | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/problems/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md index 6c04e7de..3de7c4f7 100644 --- a/problems/0035.搜索插入位置.md +++ b/problems/0035.搜索插入位置.md @@ -283,6 +283,28 @@ var searchInsert = function (nums, target) { }; ``` +### TypeScript + +```typescript +// 第一种二分法 +function searchInsert(nums: number[], target: number): number { + const length: number = nums.length; + let left: number = 0, + right: number = length - 1; + while (left <= right) { + const mid: number = Math.floor((left + right) / 2); + if (nums[mid] < target) { + left = mid + 1; + } else if (nums[mid] === target) { + return mid; + } else { + right = mid - 1; + } + } + return right + 1; +}; +``` + ### Swift ```swift From 6cd58dfef0a22f568bb5c6b03f4026520b7283de Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 4 Jun 2022 13:13:33 +0800 Subject: [PATCH 006/136] =?UTF-8?q?=E4=BF=AE=E6=94=B9=EF=BC=880024.?= =?UTF-8?q?=E4=B8=A4=E4=B8=A4=E4=BA=A4=E6=8D=A2=E9=93=BE=E8=A1=A8=E4=B8=AD?= =?UTF-8?q?=E7=9A=84=E8=8A=82=E7=82=B9.md=EF=BC=89=EF=BC=9A=E4=BC=98?= =?UTF-8?q?=E5=8C=96typescript=E7=89=88=E6=9C=AC=E4=BB=A3=E7=A0=81?= =?UTF-8?q?=EF=BC=8C=E5=A2=9E=E5=BC=BA=E6=98=93=E8=AF=BB=E6=80=A7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0024.两两交换链表中的节点.md | 27 +++++++++++++-------------- 1 file changed, 13 insertions(+), 14 deletions(-) diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md index 2289c229..0d848e4d 100644 --- a/problems/0024.两两交换链表中的节点.md +++ b/problems/0024.两两交换链表中的节点.md @@ -254,20 +254,19 @@ TypeScript： ```typescript function swapPairs(head: ListNode | null): ListNode | null { - const dummyHead: ListNode = new ListNode(0, head); - let cur: ListNode = dummyHead; - while(cur.next !== null && cur.next.next !== null) { - const tem: ListNode = cur.next; - const tem1: ListNode = cur.next.next.next; - - cur.next = cur.next.next; // step 1 - cur.next.next = tem; // step 2 - cur.next.next.next = tem1; // step 3 - - cur = cur.next.next; - } - return dummyHead.next; -} + const dummyNode: ListNode = new ListNode(0, head); + let curNode: ListNode | null = dummyNode; + while (curNode && curNode.next && curNode.next.next) { + let firstNode: ListNode = curNode.next, + secNode: ListNode = curNode.next.next, + thirdNode: ListNode | null = curNode.next.next.next; + curNode.next = secNode; + secNode.next = firstNode; + firstNode.next = thirdNode; + curNode = firstNode; + } + return dummyNode.next; +}; ``` Kotlin: From 7d5856d0b82baad2266555f33bc1eae43538fd70 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 4 Jun 2022 15:56:22 +0800 Subject: [PATCH 007/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880234.?= =?UTF-8?q?=E5=9B=9E=E6=96=87=E9=93=BE=E8=A1=A8.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0234.回文链表.md | 59 ++++++++++++++++++++++++++++++++++++++- 1 file changed, 58 insertions(+), 1 deletion(-) diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md index db910d4e..b19a2408 100644 --- a/problems/0234.回文链表.md +++ b/problems/0234.回文链表.md @@ -273,7 +273,7 @@ class Solution: return pre ``` -## Go +### Go ```go @@ -319,6 +319,63 @@ var isPalindrome = function(head) { }; ``` +### TypeScript + +> 数组模拟 + +```typescript +function isPalindrome(head: ListNode | null): boolean { + const helperArr: number[] = []; + let curNode: ListNode | null = head; + while (curNode !== null) { + helperArr.push(curNode.val); + curNode = curNode.next; + } + let left: number = 0, + right: number = helperArr.length - 1; + while (left < right) { + if (helperArr[left++] !== helperArr[right--]) return false; + } + return true; +}; +``` + +> 反转后半部分链表 + +```typescript +function isPalindrome(head: ListNode | null): boolean { + if (head === null || head.next === null) return true; + let fastNode: ListNode | null = head, + slowNode: ListNode = head, + preNode: ListNode = head; + while (fastNode !== null && fastNode.next !== null) { + preNode = slowNode; + slowNode = slowNode.next!; + fastNode = fastNode.next.next; + } + preNode.next = null; + let cur1: ListNode | null = head; + let cur2: ListNode | null = reverseList(slowNode); + while (cur1 !== null) { + if (cur1.val !== cur2!.val) return false; + cur1 = cur1.next; + cur2 = cur2!.next; + } + return true; +}; +function reverseList(head: ListNode | null): ListNode | null { + let curNode: ListNode | null = head, + preNode: ListNode | null = null; + while (curNode !== null) { + let tempNode: ListNode | null = curNode.next; + curNode.next = preNode; + preNode = curNode; + curNode = tempNode; + } + return preNode; +} +``` + ----------------------- From bb484a70ac2c8d2fcbbba08a1897bc053c9354d1 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 4 Jun 2022 21:01:59 +0800 Subject: [PATCH 008/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880143.?= =?UTF-8?q?=E9=87=8D=E6=8E=92=E9=93=BE=E8=A1=A8.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0143.重排链表.md | 76 +++++++++++++++++++++++++++++++++++++++ 1 file changed, 76 insertions(+) diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md index 790bcb48..c60fc0f9 100644 --- a/problems/0143.重排链表.md +++ b/problems/0143.重排链表.md @@ -6,6 +6,8 @@ # 143.重排链表 +[力扣题目链接](https://leetcode.cn/problems/reorder-list/submissions/) + ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20210726160122.png) ## 思路 @@ -465,7 +467,81 @@ var reorderList = function(head, s = [], tmp) { } ``` +### TypeScript + +> 辅助数组法： + +```typescript +function reorderList(head: ListNode | null): void { + if (head === null) return; + const helperArr: ListNode[] = []; + let curNode: ListNode | null = head; + while (curNode !== null) { + helperArr.push(curNode); + curNode = curNode.next; + } + let node: ListNode = head; + let left: number = 1, + right: number = helperArr.length - 1; + let count: number = 0; + while (left <= right) { + if (count % 2 === 0) { + node.next = helperArr[right--]; + } else { + node.next = helperArr[left++]; + } + count++; + node = node.next; + } + node.next = null; +}; +``` + +> 分割链表法： + +```typescript +function reorderList(head: ListNode | null): void { + if (head === null || head.next === null) return; + let fastNode: ListNode = head, + slowNode: ListNode = head; + while (fastNode.next !== null && fastNode.next.next !== null) { + slowNode = slowNode.next!; + fastNode = fastNode.next.next; + } + let head1: ListNode | null = head; + // 反转后半部分链表 + let head2: ListNode | null = reverseList(slowNode.next); + // 分割链表 + slowNode.next = null; + /** + 直接在head1链表上进行插入 + head1 链表长度一定大于或等于head2, + 因此在下面的循环中，只要head2不为null, head1 一定不为null + */ + while (head2 !== null) { + const tempNode1: ListNode | null = head1!.next, + tempNode2: ListNode | null = head2.next; + head1!.next = head2; + head2.next = tempNode1; + head1 = tempNode1; + head2 = tempNode2; + } +}; +function reverseList(head: ListNode | null): ListNode | null { + let curNode: ListNode | null = head, + preNode: ListNode | null = null; + while (curNode !== null) { + const tempNode: ListNode | null = curNode.next; + curNode.next = preNode; + preNode = curNode; + curNode = tempNode; + } + return preNode; +} +``` + ### C + 方法三：反转链表 ```c //翻转链表 From c0a73e2544202baa3839852777e053e82937004c Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 4 Jun 2022 21:35:24 +0800 Subject: [PATCH 009/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200131.=E5=88=86?= =?UTF-8?q?=E5=89=B2=E5=9B=9E=E6=96=87=E4=B8=B2.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0131.分割回文串.md | 45 +++++++++++++++++++++++++++++++++++++ 1 file changed, 45 insertions(+) diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md index 7a702898..f361d1ef 100644 --- a/problems/0131.分割回文串.md +++ b/problems/0131.分割回文串.md @@ -676,5 +676,50 @@ impl Solution { } } ``` + + +## Scala + +```scala +object Solution { + + import scala.collection.mutable + + def partition(s: String): List[List[String]] = { + var result = mutable.ListBuffer[List[String]]() + var path = mutable.ListBuffer[String]() + + // 判断字符串是否回文 + def isPalindrome(start: Int, end: Int): Boolean = { + var (left, right) = (start, end) + while (left < right) { + if (s(left) != s(right)) return false + left += 1 + right -= 1 + } + true + } + + // 回溯算法 + def backtracking(startIndex: Int): Unit = { + if (startIndex >= s.size) { + result.append(path.toList) + return + } + // 添加循环守卫，如果当前分割是回文子串则进入回溯 + for (i <- startIndex until s.size if isPalindrome(startIndex, i)) { + path.append(s.substring(startIndex, i + 1)) + backtracking(i + 1) + path = path.take(path.size - 1) + } + } + + backtracking(0) + result.toList + } +} +``` + + -----------------------
From a8e4d4da767758488dbbc20564eab01a08e3e2dc Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 4 Jun 2022 22:04:15 +0800 Subject: [PATCH 010/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200093.=E5=A4=8D?= =?UTF-8?q?=E5=8E=9FIP=E5=9C=B0=E5=9D=80.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0093.复原IP地址.md | 42 +++++++++++++++++++++++++++++++++++++ 1 file changed, 42 insertions(+) diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md index 6401824b..48bb6df2 100644 --- a/problems/0093.复原IP地址.md +++ b/problems/0093.复原IP地址.md @@ -659,6 +659,48 @@ func restoreIpAddresses(_ s: String) -> [String] { } ``` +## Scala + +```scala +object Solution { + import scala.collection.mutable + def restoreIpAddresses(s: String): List[String] = { + var result = mutable.ListBuffer[String]() + if (s.size < 4 || s.length > 12) return result.toList + var path = mutable.ListBuffer[String]() + + // 判断IP中的一个字段是否为正确的 + def isIP(sub: String): Boolean = { + if (sub.size > 1 && sub(0) == '0') return false + if (sub.toInt > 255) return false + true + } + + def backtracking(startIndex: Int): Unit = { + if (startIndex >= s.size) { + if (path.size == 4) { + result.append(path.mkString(".")) // mkString方法可以把集合里的数据以指定字符串拼接 + return + } + return + } + // subString + for (i <- startIndex until startIndex + 3 if i < s.size) { + var subString = s.substring(startIndex, i + 1) + if (isIP(subString)) { // 如果合法则进行下一轮 + path.append(subString) + backtracking(i + 1) + path = path.take(path.size - 1) + } + } + } + + backtracking(0) + result.toList + } +} +``` + -----------------------
From f5268ee213dafe8e7ab1c28d722a6b7dc81f4029 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 4 Jun 2022 22:26:58 +0800 Subject: [PATCH 011/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880141.?= =?UTF-8?q?=E7=8E=AF=E5=BD=A2=E9=93=BE=E8=A1=A8.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0141.环形链表.md | 21 ++++++++++++++++++++- 1 file changed, 20 insertions(+), 1 deletion(-) diff --git a/problems/0141.环形链表.md b/problems/0141.环形链表.md index ddd83c94..ce90b6c4 100644 --- a/problems/0141.环形链表.md +++ b/problems/0141.环形链表.md @@ -7,6 +7,8 @@ # 141. 环形链表 +[力扣题目链接](https://leetcode.cn/problems/linked-list-cycle/submissions/) + 给定一个链表，判断链表中是否有环。 如果链表中有某个节点，可以通过连续跟踪 next 指针再次到达，则链表中存在环。 为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。注意：pos 不作为参数进行传递，仅仅是为了标识链表的实际情况。 @@ -103,7 +105,7 @@ class Solution: return False ``` -## Go +### Go ```go func hasCycle(head *ListNode) bool { @@ -139,6 +141,23 @@ var hasCycle = function(head) { }; ``` +### TypeScript + +```typescript +function hasCycle(head: ListNode | null): boolean { + let slowNode: ListNode | null = head, + fastNode: ListNode | null = head; + while (fastNode !== null && fastNode.next !== null) { + slowNode = slowNode!.next; + fastNode = fastNode.next.next; + if (slowNode === fastNode) return true; + } + return false; +}; +``` + + + -----------------------
From 0245b3ae50cc612ac7765976171b670a954f6a34 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 4 Jun 2022 23:38:12 +0800 Subject: [PATCH 012/136] =?UTF-8?q?=E4=BF=AE=E6=94=B9=EF=BC=880142.?= =?UTF-8?q?=E7=8E=AF=E5=BD=A2=E9=93=BE=E8=A1=A8II.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E4=BC=98=E5=8C=96typescript=E7=89=88=E6=9C=AC=E4=BB=A3?= =?UTF-8?q?=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0142.环形链表II.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md index f8e62d45..7d3c8443 100644 --- a/problems/0142.环形链表II.md +++ b/problems/0142.环形链表II.md @@ -301,13 +301,13 @@ function detectCycle(head: ListNode | null): ListNode | null { let slowNode: ListNode | null = head, fastNode: ListNode | null = head; while (fastNode !== null && fastNode.next !== null) { - slowNode = (slowNode as ListNode).next; + slowNode = slowNode!.next; fastNode = fastNode.next.next; if (slowNode === fastNode) { slowNode = head; while (slowNode !== fastNode) { - slowNode = (slowNode as ListNode).next; - fastNode = (fastNode as ListNode).next; + slowNode = slowNode!.next; + fastNode = fastNode!.next; } return slowNode; } From bc3d202797ee1a2c276efd2b0b86cc00bd0f744a Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 5 Jun 2022 10:54:31 +0800 Subject: [PATCH 013/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880205.?= =?UTF-8?q?=E5=90=8C=E6=9E=84=E5=AD=97=E7=AC=A6=E4=B8=B2.md=EF=BC=89?= =?UTF-8?q?=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0205.同构字符串.md | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/problems/0205.同构字符串.md b/problems/0205.同构字符串.md index d4b71c59..337dcc73 100644 --- a/problems/0205.同构字符串.md +++ b/problems/0205.同构字符串.md @@ -156,6 +156,28 @@ var isIsomorphic = function(s, t) { }; ``` +## TypeScript + +```typescript +function isIsomorphic(s: string, t: string): boolean { + const helperMap1: Map = new Map(); + const helperMap2: Map = new Map(); + for (let i = 0, length = s.length; i < length; i++) { + let temp1: string | undefined = helperMap1.get(s[i]); + let temp2: string | undefined = helperMap2.get(t[i]); + if (temp1 === undefined && temp2 === undefined) { + helperMap1.set(s[i], t[i]); + helperMap2.set(t[i], s[i]); + } else if (temp1 !== t[i] || temp2 !== s[i]) { + return false; + } + } + return true; +}; +``` + + + -----------------------
From ccfb805417aa1f55efd1b30be662db8a92b46893 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sun, 5 Jun 2022 13:50:27 +0800 Subject: [PATCH 014/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200078.=E5=AD=90?= =?UTF-8?q?=E9=9B=86.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0078.子集.md | 54 +++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 54 insertions(+) diff --git a/problems/0078.子集.md b/problems/0078.子集.md index e1c52b5b..3fc396a2 100644 --- a/problems/0078.子集.md +++ b/problems/0078.子集.md @@ -373,6 +373,60 @@ func subsets(_ nums: [Int]) -> [[Int]] { } ``` +## Scala + +思路一: 使用本题解思路 + +```scala +object Solution { + import scala.collection.mutable + def subsets(nums: Array[Int]): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() + var path = mutable.ListBuffer[Int]() + + def backtracking(startIndex: Int): Unit = { + result.append(path.toList) // 存放结果 + if (startIndex >= nums.size) { + return + } + for (i <- startIndex until nums.size) { + path.append(nums(i)) // 添加元素 + backtracking(i + 1) + path.remove(path.size - 1) // 删除 + } + } + + backtracking(0) + result.toList + } +} +``` + +思路二: 将原问题转换为二叉树，针对每一个元素都有**选或不选**两种选择，直到遍历到最后，所有的叶子节点即为本题的答案： + +```scala +object Solution { + import scala.collection.mutable + def subsets(nums: Array[Int]): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() + + def backtracking(path: mutable.ListBuffer[Int], startIndex: Int): Unit = { + if (startIndex == nums.length) { + result.append(path.toList) + return + } + path.append(nums(startIndex)) + backtracking(path, startIndex + 1) // 选择元素 + path.remove(path.size - 1) + backtracking(path, startIndex + 1) // 不选择元素 + } + + backtracking(mutable.ListBuffer[Int](), 0) + result.toList + } +} +``` + -----------------------
From b3530189489235f1212b89bd9dba0281154f366c Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sun, 5 Jun 2022 14:13:41 +0800 Subject: [PATCH 015/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200090.=E5=AD=90?= =?UTF-8?q?=E9=9B=86II.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0090.子集II.md | 57 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 57 insertions(+) diff --git a/problems/0090.子集II.md b/problems/0090.子集II.md index 74ce000b..9047a809 100644 --- a/problems/0090.子集II.md +++ b/problems/0090.子集II.md @@ -434,6 +434,63 @@ func subsetsWithDup(_ nums: [Int]) -> [[Int]] { } ``` +### Scala + +不使用userd数组: + +```scala +object Solution { + import scala.collection.mutable + def subsetsWithDup(nums: Array[Int]): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() + var path = mutable.ListBuffer[Int]() + var num = nums.sorted // 排序 + + def backtracking(startIndex: Int): Unit = { + result.append(path.toList) + if (startIndex >= num.size){ + return + } + for (i <- startIndex until num.size) { + // 同一树层重复的元素不进入回溯 + if (!(i > startIndex && num(i) == num(i - 1))) { + path.append(num(i)) + backtracking(i + 1) + path.remove(path.size - 1) + } + } + } + + backtracking(0) + result.toList + } +} +``` + +使用Set去重: +```scala +object Solution { + import scala.collection.mutable + def subsetsWithDup(nums: Array[Int]): List[List[Int]] = { + var result = mutable.Set[List[Int]]() + var num = nums.sorted + def backtracking(path: mutable.ListBuffer[Int], startIndex: Int): Unit = { + if (startIndex == num.length) { + result.add(path.toList) + return + } + path.append(num(startIndex)) + backtracking(path, startIndex + 1) // 选择 + path.remove(path.size - 1) + backtracking(path, startIndex + 1) // 不选择 + } + + backtracking(mutable.ListBuffer[Int](), 0) + + result.toList + } +} +``` -----------------------
From 068cc095353ceb240c98a6c2b8598e98c445260b Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sun, 5 Jun 2022 14:46:21 +0800 Subject: [PATCH 016/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200491.=E9=80=92?= =?UTF-8?q?=E5=A2=9E=E5=AD=90=E5=BA=8F=E5=88=97.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0491.递增子序列.md | 34 ++++++++++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) diff --git a/problems/0491.递增子序列.md b/problems/0491.递增子序列.md index 3ea2382b..a04d433b 100644 --- a/problems/0491.递增子序列.md +++ b/problems/0491.递增子序列.md @@ -522,5 +522,39 @@ func findSubsequences(_ nums: [Int]) -> [[Int]] { ``` +## Scala + +```scala +object Solution { + import scala.collection.mutable + def findSubsequences(nums: Array[Int]): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() + var path = mutable.ListBuffer[Int]() + + def backtracking(startIndex: Int): Unit = { + // 集合元素大于1，添加到结果集 + if (path.size > 1) { + result.append(path.toList) + } + + var used = new Array[Boolean](201) + // 使用循环守卫，当前层没有用过的元素才有资格进入回溯 + for (i <- startIndex until nums.size if !used(nums(i) + 100)) { + // 如果path没元素或 当前循环的元素比path的最后一个元素大，则可以进入回溯 + if (path.size == 0 || (!path.isEmpty && nums(i) >= path(path.size - 1))) { + used(nums(i) + 100) = true + path.append(nums(i)) + backtracking(i + 1) + path.remove(path.size - 1) + } + } + } + + backtracking(0) + result.toList + } +} +``` + -----------------------
From 9f38af3e3a61562b17530302e8d14a6a02d1835f Mon Sep 17 00:00:00 2001 From: Hang <69448559+silaslll@users.noreply.github.com> Date: Sun, 5 Jun 2022 17:59:15 -0400 Subject: [PATCH 017/136] =?UTF-8?q?Update=200452.=E7=94=A8=E6=9C=80?= =?UTF-8?q?=E5=B0=91=E6=95=B0=E9=87=8F=E7=9A=84=E7=AE=AD=E5=BC=95=E7=88=86?= =?UTF-8?q?=E6=B0=94=E7=90=83.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新了java代码，增加了一个 leftmostRightBound variable 记录最小的右边界使得代码可读性增加 加入了comment 解释了Arrays.sort(points, (x, y) -> Integer.compare(x[0], y[0])); 中不用 x[0] - y[0] 而是用Integer.compare(x[0], y[0]) 的原因 加入了时空复杂度和说明 --- problems/0452.用最少数量的箭引爆气球.md | 21 ++++++++++++++++----- 1 file changed, 16 insertions(+), 5 deletions(-) diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md index d4bbe961..58422d4c 100644 --- a/problems/0452.用最少数量的箭引爆气球.md +++ b/problems/0452.用最少数量的箭引爆气球.md @@ -136,17 +136,28 @@ public: ### Java ```java +/** +时间复杂度 : O(NlogN) 排序需要 O(NlogN) 的复杂度 + +空间复杂度 : O(logN) java所使用的内置函数用的是快速排序需要 logN 的空间 +*/ class Solution { public int findMinArrowShots(int[][] points) { if (points.length == 0) return 0; - Arrays.sort(points, (o1, o2) -> Integer.compare(o1[0], o2[0])); - + //用x[0] - y[0] 会大于2147483647 造成整型溢出 + Arrays.sort(points, (x, y) -> Integer.compare(x[0], y[0])); + //count = 1 因为最少需要一个箭来射击第一个气球 int count = 1; - for (int i = 1; i < points.length; i++) { - if (points[i][0] > points[i - 1][1]) { + //重叠气球的最小右边界 + int leftmostRightBound = points[0][1]; + //如果下一个气球的左边界大于最小右边界 + if (points[i][0] > leftmostRightBound ) { + //增加一次射击 count++; + leftmostRightBound = points[i][1]; + //不然就更新最小右边界 } else { - points[i][1] = Math.min(points[i][1],points[i - 1][1]); + leftmostRightBound = Math.min(leftmostRightBound , points[i][1]); } } return count; From 6d804e2b244891ab3c1a12ac9b34d3eafd64654c Mon Sep 17 00:00:00 2001 From: Grant Yang Date: Sun, 5 Jun 2022 21:24:56 -0400 Subject: [PATCH 018/136] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20111.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E6=B7=B1=E5=BA=A6?= =?UTF-8?q?=20=E7=9A=84=E6=8B=BC=E5=86=99=E9=94=99=E8=AF=AF=E5=8F=8A?= =?UTF-8?q?=E6=94=B9=E8=BF=9B=E8=A7=84=E8=8C=83?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 12 ++++++------ 1 file changed, 6 insertions(+), 6 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 1743243d..d9fd0b30 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -2393,21 +2393,21 @@ JavaScript： var minDepth = function(root) { if (root === null) return 0; let queue = [root]; - let deepth = 0; + let depth = 0; while (queue.length) { let n = queue.length; - deepth++; + depth++; for (let i=0; i Date: Sun, 5 Jun 2022 22:31:41 -0400 Subject: [PATCH 019/136] =?UTF-8?q?Update=200056.=E5=90=88=E5=B9=B6?= =?UTF-8?q?=E5=8C=BA=E9=97=B4.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0056.合并区间.md | 26 ++++++++++++++++++++------ 1 file changed, 20 insertions(+), 6 deletions(-) diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index e444a221..98848963 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -136,24 +136,38 @@ public: ### Java ```java + +/** +时间复杂度 ： O(NlogN) 排序需要O(NlogN) +空间复杂度 ： O(logN) java 的内置排序是快速排序 需要 O(logN)空间 + +*/ class Solution { public int[][] merge(int[][] intervals) { List res = new LinkedList<>(); - Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[0], o2[0])); - + //按照左边界排序 + Arrays.sort(intervals, (x, y) -> Integer.compare(x[0], y[0])); + //initial start 是最小左边界 int start = intervals[0][0]; + int rightmostRightBound = intervals[0][1]; for (int i = 1; i < intervals.length; i++) { - if (intervals[i][0] > intervals[i - 1][1]) { - res.add(new int[]{start, intervals[i - 1][1]}); + //如果左边界大于最大右边界 + if (intervals[i][0] > rightmostRightBound) { + //加入区间 并且更新start + res.add(new int[]{start, rightmostRightBound}); start = intervals[i][0]; + rightmostRightBound = intervals[i][1]; } else { - intervals[i][1] = Math.max(intervals[i][1], intervals[i - 1][1]); + //更新最大右边界 + rightmostRightBound = Math.max(rightmostRightBound, intervals[i][1]); } } - res.add(new int[]{start, intervals[intervals.length - 1][1]}); + res.add(new int[]{start, rightmostRightBound}); return res.toArray(new int[res.size()][]); } } + +} ``` ```java // 版本2 From dd00b57be87e9c4e25267c5ab6cc53f6d714a97e Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Mon, 6 Jun 2022 21:43:18 -0500 Subject: [PATCH 020/136] =?UTF-8?q?Update=200704.=E4=BA=8C=E5=88=86?= =?UTF-8?q?=E6=9F=A5=E6=89=BE.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0704.二分查找.md | 24 +++++++++++++----------- 1 file changed, 13 insertions(+), 11 deletions(-) diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index 1e474f9a..a468cc44 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -218,19 +218,21 @@ class Solution: （版本二）左闭右开区间 -```python -class Solution: +```class Solution: def search(self, nums: List[int], target: int) -> int: - left,right =0, len(nums) - while left < right: - mid = (left + right) // 2 - if nums[mid] < target: - left = mid+1 - elif nums[mid] > target: - right = mid + if nums is None or len(nums)==0: + return -1 + l=0 + r=len(nums)-1 + while (l<=r): + m = round(l+(r-l)/2) + if nums[m] == target: + return m + elif nums[m] > target: + r=m-1 else: - return mid - return -1 + l=m+1 + return -1 ``` **Go：** From 091204c926a044f0ec86200cbdd3cca6deeaf97e Mon Sep 17 00:00:00 2001 From: HanMengnan <1448189829@qq.com> Date: Tue, 7 Jun 2022 10:46:43 +0800 Subject: [PATCH 021/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880129.?= =?UTF-8?q?=E6=B1=82=E6=A0=B9=E8=8A=82=E7=82=B9=E5=88=B0=E5=8F=B6=E8=8A=82?= =?UTF-8?q?=E7=82=B9=E6=95=B0=E5=AD=97=E4=B9=8B=E5=92=8C.md=EF=BC=89?= =?UTF-8?q?=EF=BC=9A=E5=A2=9E=E5=8A=A0go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0129.求根到叶子节点数字之和.md | 26 +++++++++++++++++++++++++ 1 file changed, 26 insertions(+) diff --git a/problems/0129.求根到叶子节点数字之和.md b/problems/0129.求根到叶子节点数字之和.md index b271ca7d..4191bb26 100644 --- a/problems/0129.求根到叶子节点数字之和.md +++ b/problems/0129.求根到叶子节点数字之和.md @@ -3,6 +3,9 @@

参与本项目，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！

+ + + # 129. 求根节点到叶节点数字之和 [力扣题目链接](https://leetcode-cn.com/problems/sum-root-to-leaf-numbers/) @@ -245,6 +248,29 @@ class Solution: ``` Go： +```go +func sumNumbers(root *TreeNode) int { + sum = 0 + travel(root, root.Val) + return sum +} + +func travel(root *TreeNode, tmpSum int) { + if root.Left == nil && root.Right == nil { + sum += tmpSum + } else { + if root.Left != nil { + travel(root.Left, tmpSum*10+root.Left.Val) + } + if root.Right != nil { + travel(root.Right, tmpSum*10+root.Right.Val) + } + } +} +``` + + + JavaScript： ```javascript var sumNumbers = function(root) { From f49b2e4a75c6f98c082923e1098aedc5cb27574b Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Mon, 6 Jun 2022 21:49:32 -0500 Subject: [PATCH 022/136] =?UTF-8?q?Update=200027.=E7=A7=BB=E9=99=A4?= =?UTF-8?q?=E5=85=83=E7=B4=A0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit the former code has not consider if nums is None or len(nums)==0 --- problems/0027.移除元素.md | 38 +++++++++++++++++--------------------- 1 file changed, 17 insertions(+), 21 deletions(-) diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 4b50d666..b239136c 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -173,28 +173,24 @@ class Solution { Python： -```python +```python3 class Solution: - """双指针法 - 时间复杂度：O(n) - 空间复杂度：O(1) - """ - - @classmethod - def removeElement(cls, nums: List[int], val: int) -> int: - fast = slow = 0 - - while fast < len(nums): - - if nums[fast] != val: - nums[slow] = nums[fast] - slow += 1 - - # 当 fast 指针遇到要删除的元素时停止赋值 - # slow 指针停止移动, fast 指针继续前进 - fast += 1 - - return slow + def removeElement(self, nums: List[int], val: int) -> int: + if nums is None or len(nums)==0: + return 0 + l=0 + r=len(nums)-1 + while l Date: Mon, 6 Jun 2022 22:11:16 -0500 Subject: [PATCH 023/136] =?UTF-8?q?Update=200977.=E6=9C=89=E5=BA=8F?= =?UTF-8?q?=E6=95=B0=E7=BB=84=E7=9A=84=E5=B9=B3=E6=96=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit python3 version of brutal force --- problems/0977.有序数组的平方.md | 9 +++++++++ 1 file changed, 9 insertions(+) diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index 0e79a3d6..d3da662f 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -39,6 +39,15 @@ public: } }; ``` +```python3 +class Solution: + def sortedSquares(self, nums: List[int]) -> List[int]: + res=[] + for num in nums: + res.append(num**2) + return sorted(res) +``` + 这个时间复杂度是 O(n + nlogn)， 可以说是O(nlogn)的时间复杂度，但为了和下面双指针法算法时间复杂度有鲜明对比，我记为 O(n + nlog n)。 From b10f7edba037886e736e58bbe0488ab078efdaa2 Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Mon, 6 Jun 2022 22:17:15 -0500 Subject: [PATCH 024/136] =?UTF-8?q?Update=200209.=E9=95=BF=E5=BA=A6?= =?UTF-8?q?=E6=9C=80=E5=B0=8F=E7=9A=84=E5=AD=90=E6=95=B0=E7=BB=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 滑动窗口 version of python3 code --- problems/0209.长度最小的子数组.md | 23 +++++++++++++++++++++-- 1 file changed, 21 insertions(+), 2 deletions(-) diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index fbef7692..160f93bb 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -162,8 +162,27 @@ class Solution: index += 1 return 0 if res==float("inf") else res ``` - - +```python3 +#滑动窗口 +class Solution: + def minSubArrayLen(self, target: int, nums: List[int]) -> int: + if nums is None or len(nums)==0: + return 0 + lenf=len(nums)+1 + total=0 + i=j=0 + while (j=target): + lenf=min(lenf,j-i) + total=total-nums[i] + i+=1 + if lenf==len(nums)+1: + return 0 + else: + return lenf +``` Go： ```go func minSubArrayLen(target int, nums []int) int { From 97fc88e533418bf9070bd9fb549a23a2499805b4 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Tue, 7 Jun 2022 16:59:55 +0800 Subject: [PATCH 025/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200046.=E5=85=A8?= =?UTF-8?q?=E6=8E=92=E5=88=97.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0046.全排列.md | 30 ++++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md index 836c3646..ec3adaa7 100644 --- a/problems/0046.全排列.md +++ b/problems/0046.全排列.md @@ -456,6 +456,36 @@ func permute(_ nums: [Int]) -> [[Int]] { } ``` +### Scala + +```scala +object Solution { + import scala.collection.mutable + def permute(nums: Array[Int]): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() + var path = mutable.ListBuffer[Int]() + + def backtracking(used: Array[Boolean]): Unit = { + if (path.size == nums.size) { + // 如果path的长度和nums相等，那么可以添加到结果集 + result.append(path.toList) + return + } + // 添加循环守卫，只有当当前数字没有用过的情况下才进入回溯 + for (i <- nums.indices if used(i) == false) { + used(i) = true + path.append(nums(i)) + backtracking(used) // 回溯 + path.remove(path.size - 1) + used(i) = false + } + } + + backtracking(new Array[Boolean](nums.size)) // 调用方法 + result.toList // 最终返回结果集的List形式 + } +} +``` -----------------------
From 8537401616f10ce85038bbd94aa951a20bb877d6 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Tue, 7 Jun 2022 17:24:41 +0800 Subject: [PATCH 026/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200047.=E5=85=A8?= =?UTF-8?q?=E6=8E=92=E5=88=97II.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0047.全排列II.md | 38 ++++++++++++++++++++++++++++++++++++++ 1 file changed, 38 insertions(+) diff --git a/problems/0047.全排列II.md b/problems/0047.全排列II.md index cce25cd9..0a5debcc 100644 --- a/problems/0047.全排列II.md +++ b/problems/0047.全排列II.md @@ -422,5 +422,43 @@ int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumn } ``` +### Scala + +```scala +object Solution { + import scala.collection.mutable + def permuteUnique(nums: Array[Int]): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() + var path = mutable.ListBuffer[Int]() + var num = nums.sorted // 首先对数据进行排序 + + def backtracking(used: Array[Boolean]): Unit = { + if (path.size == num.size) { + // 如果path的size等于num了，那么可以添加到结果集 + result.append(path.toList) + return + } + // 循环守卫，当前元素没被使用过就进入循环体 + for (i <- num.indices if used(i) == false) { + // 当前索引为0，不存在和前一个数字相等可以进入回溯 + // 当前索引值和上一个索引不相等，可以回溯 + // 前一个索引对应的值没有被选，可以回溯 + // 因为Scala没有continue，只能将逻辑反过来写 + if (i == 0 || (i > 0 && num(i) != num(i - 1)) || used(i-1) == false) { + used(i) = true + path.append(num(i)) + backtracking(used) + path.remove(path.size - 1) + used(i) = false + } + } + } + + backtracking(new Array[Boolean](nums.length)) + result.toList + } +} +``` + -----------------------
From fa26fb332b43cbeec805944f4522472953232487 Mon Sep 17 00:00:00 2001 From: dcj_hp <294487055@qq.com> Date: Tue, 7 Jun 2022 17:38:37 +0800 Subject: [PATCH 027/136] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=20java=20dp=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0132.分割回文串II.md | 50 +++++++++++++++++++++++++++++++++++ 1 file changed, 50 insertions(+) diff --git a/problems/0132.分割回文串II.md b/problems/0132.分割回文串II.md index 87d3e4b4..4cb95901 100644 --- a/problems/0132.分割回文串II.md +++ b/problems/0132.分割回文串II.md @@ -206,6 +206,55 @@ public: ## Java ```java +class Solution { + + public int minCut(String s) { + if(null == s || "".equals(s)){ + return 0; + } + int len = s.length(); + // 1. + // 记录子串[i..j]是否是回文串 + boolean[][] isPalindromic = new boolean[len][len]; + // 从下到上，从左到右 + for(int i = len - 1; i >= 0; i--){ + for(int j = i; j < len; j++){ + if(s.charAt(i) == s.charAt(j)){ + if(j - i <= 1){ + isPalindromic[i][j] = true; + } else{ + isPalindromic[i][j] = isPalindromic[i + 1][j - 1]; + } + } else{ + isPalindromic[i][j] = false; + } + } + } + + // 2. + // dp[i] 表示[0..i]的最小分割次数 + int[] dp = new int[len]; + for(int i = 0; i < len; i++){ + //初始考虑最坏的情况。 1个字符分割0次， len个字符分割 len - 1次 + dp[i] = i; + } + + for(int i = 1; i < len; i++){ + if(isPalindromic[0][i]){ + // s[0..i]是回文了，那 dp[i] = 0, 一次也不用分割 + dp[i] = 0; + continue; + } + for(int j = 0; j < i; j++){ + // 按文中的思路，不清楚就拿 "ababa" 为例，先写出 isPalindromic 数组，再进行求解 + if(isPalindromic[j + 1][i]){ + dp[i] = Math.min(dp[i], dp[j] + 1); + } + } + } + return dp[len - 1]; + } +} ``` ## Python @@ -240,6 +289,7 @@ class Solution: ## Go ```go + ``` ## JavaScript From 7b95f173683f4eae7fd68c313839a678b2e1b79d Mon Sep 17 00:00:00 2001 From: Chris Chen Date: Tue, 7 Jun 2022 13:21:32 +0100 Subject: [PATCH 028/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0(0001.=E4=B8=A4?= =?UTF-8?q?=E6=95=B0=E4=B9=8B=E5=92=8C.md=EF=BC=89=EF=BC=9ADart=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0001.两数之和.md | 14 ++++++++++++++ 1 file changed, 14 insertions(+) diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index 6969c2e2..4ba92092 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -317,6 +317,20 @@ public class Solution { } ``` +Dart: +```dart +List twoSum(List nums, int target) { + var tmp = []; + for (var i = 0; i < nums.length; i++) { + var rest = target - nums[i]; + if(tmp.contains(rest)){ + return [tmp.indexOf(rest), i]; + } + tmp.add(nums[i]); + } + return [0 , 0]; +} +``` -----------------------
From 371564ba3b9a2b9641cafc05eb033f8d162aec81 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Tue, 7 Jun 2022 22:19:13 +0800 Subject: [PATCH 029/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200051.N=E7=9A=87?= =?UTF-8?q?=E5=90=8E.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0051.N皇后.md | 74 +++++++++++++++++++++++++++++++++++++++++- 1 file changed, 73 insertions(+), 1 deletion(-) diff --git a/problems/0051.N皇后.md b/problems/0051.N皇后.md index c03e48c2..f65ccaf5 100644 --- a/problems/0051.N皇后.md +++ b/problems/0051.N皇后.md @@ -455,7 +455,7 @@ var solveNQueens = function(n) { }; ``` -## TypeScript +### TypeScript ```typescript function solveNQueens(n: number): string[][] { @@ -683,5 +683,77 @@ char *** solveNQueens(int n, int* returnSize, int** returnColumnSizes){ } ``` +### Scala + +```scala +object Solution { + import scala.collection.mutable + def solveNQueens(n: Int): List[List[String]] = { + var result = mutable.ListBuffer[List[String]]() + + def judge(x: Int, y: Int, maze: Array[Array[Boolean]]): Boolean = { + // 正上方 + var xx = x + while (xx >= 0) { + if (maze(xx)(y)) return false + xx -= 1 + } + // 左边 + var yy = y + while (yy >= 0) { + if (maze(x)(yy)) return false + yy -= 1 + } + // 左上方 + xx = x + yy = y + while (xx >= 0 && yy >= 0) { + if (maze(xx)(yy)) return false + xx -= 1 + yy -= 1 + } + xx = x + yy = y + // 右上方 + while (xx >= 0 && yy < n) { + if (maze(xx)(yy)) return false + xx -= 1 + yy += 1 + } + true + } + + def backtracking(row: Int, maze: Array[Array[Boolean]]): Unit = { + if (row == n) { + // 将结果转换为题目所需要的形式 + var path = mutable.ListBuffer[String]() + for (x <- maze) { + var tmp = mutable.ListBuffer[String]() + for (y <- x) { + if (y == true) tmp.append("Q") + else tmp.append(".") + } + path.append(tmp.mkString) + } + result.append(path.toList) + return + } + + for (j <- 0 until n) { + // 判断这个位置是否可以放置皇后 + if (judge(row, j, maze)) { + maze(row)(j) = true + backtracking(row + 1, maze) + maze(row)(j) = false + } + } + } + + backtracking(0, Array.ofDim[Boolean](n, n)) + result.toList + } +} +``` + -----------------------
From b61afe9aee786592c313afb42b07295d0c1f4e00 Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Tue, 7 Jun 2022 12:30:18 -0500 Subject: [PATCH 030/136] =?UTF-8?q?Update=200054.=E8=9E=BA=E6=97=8B?= =?UTF-8?q?=E7=9F=A9=E9=98=B5.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit for loop version of python3 solution --- problems/0054.螺旋矩阵.md | 26 +++++++++++++++++++++++++- 1 file changed, 25 insertions(+), 1 deletion(-) diff --git a/problems/0054.螺旋矩阵.md b/problems/0054.螺旋矩阵.md index ccf6f471..27899d51 100644 --- a/problems/0054.螺旋矩阵.md +++ b/problems/0054.螺旋矩阵.md @@ -171,6 +171,30 @@ class Solution: return res ``` - +```python3 +class Solution: + def spiralOrder(self, matrix: List[List[int]]) -> List[int]: + r=len(matrix) + if r == 0 or len(matrix[0])==0: + return [] + c=len(matrix[0]) + res=matrix[0] + + if r>1: + for i in range (1,r): + res.append(matrix[i][c-1]) + for j in range(c-2, -1, -1): + res.append(matrix[r-1][j]) + if c>1: + for i in range(r-2, 0, -1): + res.append(matrix[i][0]) + + M=[] + for k in range(1, r-1): + e=matrix[k][1:-1] + M.append(e) + + return res+self.spiralOrder(M) +``` -----------------------
From 237dc9d6b3c648d4a76e905bbc7059c2f573e91a Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Wed, 8 Jun 2022 12:52:55 +0800 Subject: [PATCH 031/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880925.?= =?UTF-8?q?=E9=95=BF=E6=8C=89=E9=94=AE=E5=85=A5.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0925.长按键入.md | 25 +++++++++++++++++++++++++ 1 file changed, 25 insertions(+) diff --git a/problems/0925.长按键入.md b/problems/0925.长按键入.md index 0ef5a3d7..feb57391 100644 --- a/problems/0925.长按键入.md +++ b/problems/0925.长按键入.md @@ -209,6 +209,31 @@ var isLongPressedName = function(name, typed) { }; ``` +### TypeScript + +```typescript +function isLongPressedName(name: string, typed: string): boolean { + const nameLength: number = name.length, + typeLength: number = typed.length; + let i: number = 0, + j: number = 0; + while (i < nameLength && j < typeLength) { + if (name[i] !== typed[j]) return false; + i++; + j++; + if (i === nameLength || name[i] !== name[i - 1]) { + // 跳过typed中的连续相同字符 + while (j < typeLength && typed[j] === typed[j - 1]) { + j++; + } + } + } + return i === nameLength && j === typeLength; +}; +``` + + + -----------------------
From cded6c5c803e5d6de7a47e020e94ce03fcc68432 Mon Sep 17 00:00:00 2001 From: ExplosiveBattery <641370196@qq.com> Date: Thu, 9 Jun 2022 00:58:38 +0800 Subject: [PATCH 032/136] =?UTF-8?q?Update=200110.=E5=B9=B3=E8=A1=A1?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md=20=20python=20code=20via=20itera?= =?UTF-8?q?te?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit In the original method, we need to traversal every node and write the function named getDepth to get the depth of all sub trees in traverse method too. But there is more suitable uniform iteration traversal algorithm, I use the map struct in the code segment where the node is Null. If you have problem in understand, please feel free to communicate with me. --- problems/0110.平衡二叉树.md | 46 +++++++++++++------------------------ 1 file changed, 16 insertions(+), 30 deletions(-) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index d98ff8a9..1b997643 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -531,40 +531,26 @@ class Solution: 迭代法： ```python class Solution: - def isBalanced(self, root: TreeNode) -> bool: - st = [] + def isBalanced(self, root: Optional[TreeNode]) -> bool: if not root: return True - st.append(root) - while st: - node = st.pop() #中 - if abs(self.getDepth(node.left) - self.getDepth(node.right)) > 1: - return False - if node.right: - st.append(node.right) #右（空节点不入栈） - if node.left: - st.append(node.left) #左（空节点不入栈） - return True - - def getDepth(self, cur): - st = [] - if cur: - st.append(cur) - depth = 0 - result = 0 - while st: - node = st.pop() + + height_map = {} + stack = [root] + while stack: + node = stack.pop() if node: - st.append(node) #中 - st.append(None) - depth += 1 - if node.right: st.append(node.right) #右 - if node.left: st.append(node.left) #左 + stack.append(node) + stack.append(None) + if node.left: stack.append(node.left) + if node.right: stack.append(node.right) else: - node = st.pop() - depth -= 1 - result = max(result, depth) - return result + real_node = stack.pop() + left, right = height_map.get(real_node.left, 0), height_map.get(real_node.right, 0) + if abs(left - right) > 1: + return False + height_map[real_node] = 1 + max(left, right) + return True ``` From e354cd6e25231fc9b255ae1d6ed3d1c58e4aa27d Mon Sep 17 00:00:00 2001 From: ExplosiveBattery <641370196@qq.com> Date: Thu, 9 Jun 2022 02:33:55 +0800 Subject: [PATCH 033/136] =?UTF-8?q?Update=200101.=E5=AF=B9=E7=A7=B0?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md=20level=20order=20traversal?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit This leetcode problem can use level order traversal method, the difference between with the normal version is we should judge for None. There is my python answer, please feel free to contact with me if you have any problem. --- problems/0101.对称二叉树.md | 25 +++++++++++++++++++++++++ 1 file changed, 25 insertions(+) diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index 1eb43589..caf5d249 100644 --- a/problems/0101.对称二叉树.md +++ b/problems/0101.对称二叉树.md @@ -437,6 +437,31 @@ class Solution: return True ``` +层次遍历 +```python +class Solution: + def isSymmetric(self, root: Optional[TreeNode]) -> bool: + if not root: + return True + + que = [root] + while que: + this_level_length = len(que) + for i in range(this_level_length // 2): + # 要么其中一个是None但另外一个不是 + if (not que[i] and que[this_level_length - 1 - i]) or (que[i] and not que[this_level_length - 1 - i]): + return False + # 要么两个都不是None + if que[i] and que[i].val != que[this_level_length - 1 - i].val: + return False + for i in range(this_level_length): + if not que[i]: continue + que.append(que[i].left) + que.append(que[i].right) + que = que[this_level_length:] + return True +``` + ## Go ```go From ba31161609d4131a5d039046435a8f21805e301d Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Wed, 8 Jun 2022 17:47:49 -0500 Subject: [PATCH 034/136] =?UTF-8?q?Update=20=E9=9D=A2=E8=AF=95=E9=A2=9802.?= =?UTF-8?q?07.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/面试题02.07.链表相交.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 0a38cc33..ba6631a4 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -160,6 +160,8 @@ class Solution: 那么，只要其中一个链表走完了，就去走另一条链表的路。如果有交点，他们最终一定会在同一个 位置相遇 """ + if headA is None or headB is None: + return None cur_a, cur_b = headA, headB # 用两个指针代替a和b From e06b53a88d8b26b4840428255b0e1f65f0eaf938 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Thu, 9 Jun 2022 21:57:15 +0800 Subject: [PATCH 035/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200037.=E8=A7=A3?= =?UTF-8?q?=E6=95=B0=E7=8B=AC.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0037.解数独.md | 95 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 95 insertions(+) diff --git a/problems/0037.解数独.md b/problems/0037.解数独.md index c1ac15af..e2c533a9 100644 --- a/problems/0037.解数独.md +++ b/problems/0037.解数独.md @@ -602,5 +602,100 @@ func solveSudoku(_ board: inout [[Character]]) { } ``` +### Scala + +详细写法： +```scala +object Solution { + + def solveSudoku(board: Array[Array[Char]]): Unit = { + backtracking(board) + } + + def backtracking(board: Array[Array[Char]]): Boolean = { + for (i <- 0 until 9) { + for (j <- 0 until 9) { + if (board(i)(j) == '.') { // 必须是为 . 的数字才放数字 + for (k <- '1' to '9') { // 这个位置放k是否合适 + if (isVaild(i, j, k, board)) { + board(i)(j) = k + if (backtracking(board)) return true // 找到了立刻返回 + board(i)(j) = '.' // 回溯 + } + } + return false // 9个数都试完了，都不行就返回false + } + } + } + true // 遍历完所有的都没返回false，说明找到了 + } + + def isVaild(x: Int, y: Int, value: Char, board: Array[Array[Char]]): Boolean = { + // 行 + for (i <- 0 until 9 ) { + if (board(i)(y) == value) { + return false + } + } + + // 列 + for (j <- 0 until 9) { + if (board(x)(j) == value) { + return false + } + } + + // 宫 + var row = (x / 3) * 3 + var col = (y / 3) * 3 + for (i <- row until row + 3) { + for (j <- col until col + 3) { + if (board(i)(j) == value) { + return false + } + } + } + + true + } +} +``` + +遵循Scala至简原则写法： +```scala +object Solution { + + def solveSudoku(board: Array[Array[Char]]): Unit = { + backtracking(board) + } + + def backtracking(board: Array[Array[Char]]): Boolean = { + // 双重for循环 + 循环守卫 + for (i <- 0 until 9; j <- 0 until 9 if board(i)(j) == '.') { + // 必须是为 . 的数字才放数字，使用循环守卫判断该位置是否可以放置当前循环的数字 + for (k <- '1' to '9' if isVaild(i, j, k, board)) { // 这个位置放k是否合适 + board(i)(j) = k + if (backtracking(board)) return true // 找到了立刻返回 + board(i)(j) = '.' // 回溯 + } + return false // 9个数都试完了，都不行就返回false + } + true // 遍历完所有的都没返回false，说明找到了 + } + + def isVaild(x: Int, y: Int, value: Char, board: Array[Array[Char]]): Boolean = { + // 行，循环守卫进行判断 + for (i <- 0 until 9 if board(i)(y) == value) return false + // 列，循环守卫进行判断 + for (j <- 0 until 9 if board(x)(j) == value) return false + // 宫，循环守卫进行判断 + var row = (x / 3) * 3 + var col = (y / 3) * 3 + for (i <- row until row + 3; j <- col until col + 3 if board(i)(j) == value) return false + true // 最终没有返回false，就说明该位置可以填写true + } +} +``` + -----------------------
From 730f58bccee54b576c2b5bc998e5b97fe4fb5567 Mon Sep 17 00:00:00 2001 From: hongyang <1664698982@qq.com> Date: Thu, 9 Jun 2022 23:24:30 +0800 Subject: [PATCH 036/136] refactor: add golang solution to Intersection of Two Linked Lists LCCI --- problems/面试题02.07.链表相交.md | 53 +++++++++++++++++++++++--------- 1 file changed, 38 insertions(+), 15 deletions(-) diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 0a38cc33..f603925d 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -13,21 +13,21 @@ 图示两个链表在节点 c1 开始相交： -![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221657.png) +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221657.png) 题目数据 保证 整个链式结构中不存在环。 -注意，函数返回结果后，链表必须 保持其原始结构 。 +注意，函数返回结果后，链表必须 保持其原始结构 。 -示例 1： +示例 1： -![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221723.png) +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221723.png) 示例 2： -![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221749.png) +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221749.png) -示例 3： +示例 3： ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221812.png)![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221812.png) @@ -100,7 +100,7 @@ public: ## 其他语言版本 -### Java +### Java ```Java public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { @@ -144,11 +144,11 @@ public class Solution { } return null; } - + } ``` -### Python +### Python ```python class Solution: @@ -162,15 +162,15 @@ class Solution: """ cur_a, cur_b = headA, headB # 用两个指针代替a和b - + while cur_a != cur_b: cur_a = cur_a.next if cur_a else headB # 如果a走完了，那么就切换到b走 cur_b = cur_b.next if cur_b else headA # 同理，b走完了就切换到a - + return cur_a ``` -### Go +### Go ```go func getIntersectionNode(headA, headB *ListNode) *ListNode { @@ -208,7 +208,30 @@ func getIntersectionNode(headA, headB *ListNode) *ListNode { } ``` -### javaScript +递归版本： + +```go +func getIntersectionNode(headA, headB *ListNode) *ListNode { + l1,l2 := headA, headB + for l1 != l2 { + if l1 != nil { + l1 = l1.Next + } else { + l1 = headB + } + + if l2 != nil { + l2 = l2.Next + } else { + l2 = headA + } + } + + return l1 +} +``` + +### javaScript ```js var getListLen = function(head) { @@ -218,9 +241,9 @@ var getListLen = function(head) { cur = cur.next; } return len; -} +} var getIntersectionNode = function(headA, headB) { - let curA = headA,curB = headB, + let curA = headA,curB = headB, lenA = getListLen(headA), lenB = getListLen(headB); if(lenA < lenB) { From c9900267505297634675172ad670e86a2269cace Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Thu, 9 Jun 2022 21:11:08 -0500 Subject: [PATCH 037/136] =?UTF-8?q?Update=200234.=E5=9B=9E=E6=96=87?= =?UTF-8?q?=E9=93=BE=E8=A1=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit For both solution of python3, there are shorter and more efficient ways to write it. For the #数组模拟, it can be solved more easily by convert the linked list to a list #反转后半部分链表, the original version define to function, isPalindrome, and reverseList. That's too complicated... No need. --- problems/0234.回文链表.md | 80 +++++++++++++++------------------------ 1 file changed, 31 insertions(+), 49 deletions(-) diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md index db910d4e..bbfe4e91 100644 --- a/problems/0234.回文链表.md +++ b/problems/0234.回文链表.md @@ -218,59 +218,41 @@ class Solution { ```python #数组模拟 class Solution: - def isPalindrome(self, head: ListNode) -> bool: - length = 0 - tmp = head - while tmp: #求链表长度 - length += 1 - tmp = tmp.next - - result = [0] * length - tmp = head - index = 0 - while tmp: #链表元素加入数组 - result[index] = tmp.val - index += 1 - tmp = tmp.next - - i, j = 0, length - 1 - while i < j: # 判断回文 - if result[i] != result[j]: + def isPalindrome(self, head: Optional[ListNode]) -> bool: + list=[] + while head: + list.append(head.val) + head=head.next + l,r=0, len(list)-1 + while l<=r: + if list[l]!=list[r]: return False - i += 1 - j -= 1 - return True - + l+=1 + r-=1 + return True + #反转后半部分链表 class Solution: - def isPalindrome(self, head: ListNode) -> bool: - if head == None or head.next == None: - return True - slow, fast = head, head - while fast and fast.next: - pre = slow - slow = slow.next - fast = fast.next.next - - pre.next = None # 分割链表 - cur1 = head # 前半部分 - cur2 = self.reverseList(slow) # 反转后半部分，总链表长度如果是奇数，cur2比cur1多一个节点 - while cur1: - if cur1.val != cur2.val: - return False - cur1 = cur1.next - cur2 = cur2.next - return True + def isPalindrome(self, head: Optional[ListNode]) -> bool: + fast = slow = head - def reverseList(self, head: ListNode) -> ListNode: - cur = head - pre = None - while(cur!=None): - temp = cur.next # 保存一下cur的下一个节点 - cur.next = pre # 反转 - pre = cur - cur = temp - return pre + # find mid point which including (first) mid point into the first half linked list + while fast and fast.next: + fast = fast.next.next + slow = slow.next + node = None + + # reverse second half linked list + while slow: + slow.next, slow, node = node, slow.next, slow + + # compare reversed and original half; must maintain reversed linked list is shorter than 1st half + while node: + if node.val != head.val: + return False + node = node.next + head = head.next + return True ``` ## Go From ddcd31b2a7668214e94763709d2cb1c88a731dae Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Fri, 10 Jun 2022 11:27:26 +0800 Subject: [PATCH 038/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880844.?= =?UTF-8?q?=E6=AF=94=E8=BE=83=E5=90=AB=E9=80=80=E6=A0=BC=E7=9A=84=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typesc?= =?UTF-8?q?ript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0844.比较含退格的字符串.md | 65 +++++++++++++++++++++++++++++ 1 file changed, 65 insertions(+) diff --git a/problems/0844.比较含退格的字符串.md b/problems/0844.比较含退格的字符串.md index dccc5404..017d4cef 100644 --- a/problems/0844.比较含退格的字符串.md +++ b/problems/0844.比较含退格的字符串.md @@ -399,6 +399,71 @@ var backspaceCompare = function(s, t) { ``` +### TypeScript + +> 双栈法： + +```typescript +function backspaceCompare(s: string, t: string): boolean { + const stack1: string[] = [], + stack2: string[] = []; + for (let c of s) { + if (c === '#') { + stack1.pop(); + } else { + stack1.push(c); + } + } + for (let c of t) { + if (c === '#') { + stack2.pop(); + } else { + stack2.push(c); + } + } + if (stack1.length !== stack2.length) return false; + for (let i = 0, length = stack1.length; i < length; i++) { + if (stack1[i] !== stack2[i]) return false; + } + return true; +}; +``` + +> 双指针法： + +```typescript +function backspaceCompare(s: string, t: string): boolean { + let sIndex: number = s.length - 1, + tIndex: number = t.length - 1; + while (true) { + sIndex = getIndexAfterDel(s, sIndex); + tIndex = getIndexAfterDel(t, tIndex); + if (sIndex < 0 || tIndex < 0) break; + if (s[sIndex] !== t[tIndex]) return false; + sIndex--; + tIndex--; + } + return sIndex === -1 && tIndex === -1; +}; +function getIndexAfterDel(s: string, startIndex: number): number { + let backspaceNum: number = 0; + while (startIndex >= 0) { + // 不可消除 + if (s[startIndex] !== '#' && backspaceNum === 0) break; + // 可消除 + if (s[startIndex] === '#') { + backspaceNum++; + } else { + backspaceNum--; + } + startIndex--; + } + return startIndex; +} +``` + + + -----------------------
From 18f29ef1786e41fba506bc27f35c6a83a116ba14 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Fri, 10 Jun 2022 16:28:33 +0800 Subject: [PATCH 039/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200455.=E5=88=86?= =?UTF-8?q?=E5=8F=91=E9=A5=BC=E5=B9=B2.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0455.分发饼干.md | 21 +++++++++++++++++++++ 1 file changed, 21 insertions(+) diff --git a/problems/0455.分发饼干.md b/problems/0455.分发饼干.md index 17db4a85..443ab6d7 100644 --- a/problems/0455.分发饼干.md +++ b/problems/0455.分发饼干.md @@ -296,5 +296,26 @@ int findContentChildren(int* g, int gSize, int* s, int sSize){ } ``` +### Scala + +```scala +object Solution { + def findContentChildren(g: Array[Int], s: Array[Int]): Int = { + var result = 0 + var children = g.sorted + var cookie = s.sorted + // 遍历饼干 + var j = 0 + for (i <- cookie.indices) { + if (j < children.size && cookie(i) >= children(j)) { + j += 1 + result += 1 + } + } + result + } +} +``` + -----------------------
From 7fd311f47bf257787caeea3bcbf73ec3ce29e978 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Fri, 10 Jun 2022 20:09:14 +0800 Subject: [PATCH 040/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200376.=E6=91=86?= =?UTF-8?q?=E5=8A=A8=E5=BA=8F=E5=88=97.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0376.摆动序列.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index 6822896e..fb4d6eff 100644 --- a/problems/0376.摆动序列.md +++ b/problems/0376.摆动序列.md @@ -375,7 +375,31 @@ function wiggleMaxLength(nums: number[]): number { }; ``` +### Scala +```scala +object Solution { + def wiggleMaxLength(nums: Array[Int]): Int = { + if (nums.length <= 1) return nums.length + var result = 1 + var curDiff = 0 // 当前一对的差值 + var preDiff = 0 // 前一对的差值 + + for (i <- 1 until nums.length) { + curDiff = nums(i) - nums(i - 1) // 计算当前这一对的差值 + // 当 curDiff > 0 的情况，preDiff <= 0 + // 当 curDiff < 0 的情况，preDiff >= 0 + // 这两种情况算是两个峰值 + if ((curDiff > 0 && preDiff <= 0) || (curDiff < 0 && preDiff >= 0)) { + result += 1 // 结果集加 1 + preDiff = curDiff // 当前差值赋值给上一轮 + } + } + + result + } +} +``` -----------------------
From 7ee790f4f532e56664642163a4612c7242f0fe11 Mon Sep 17 00:00:00 2001 From: ccchooko <648646891@qq.com> Date: Fri, 10 Jun 2022 22:34:11 +0800 Subject: [PATCH 041/136] =?UTF-8?q?0042=20=E6=8E=A5=E9=9B=A8=E6=B0=B4?= =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E5=8D=95=E8=B0=83=E6=A0=88go=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0042.接雨水.md | 36 ++++++++++++++++++++++++++++++++++++ 1 file changed, 36 insertions(+) diff --git a/problems/0042.接雨水.md b/problems/0042.接雨水.md index b232ce22..e331967f 100644 --- a/problems/0042.接雨水.md +++ b/problems/0042.接雨水.md @@ -640,8 +640,44 @@ func min(a,b int)int{ } ``` +单调栈解法 +```go +func trap(height []int) int { + if len(height) <= 2 { + return 0 + } + st := make([]int, 1, len(height)) // 切片模拟单调栈，st存储的是高度数组下标 + var res int + for i := 1; i < len(height); i++ { + if height[i] < height[st[len(st)-1]] { + st = append(st, i) + } else if height[i] == height[st[len(st)-1]] { + st = st[:len(st)-1] // 比较的新元素和栈顶的元素相等，去掉栈中的，入栈新元素下标 + st = append(st, i) + } else { + for len(st) != 0 && height[i] > height[st[len(st)-1]] { + top := st[len(st)-1] + st = st[:len(st)-1] + if len(st) != 0 { + tmp := (min(height[i], height[st[len(st)-1]]) - height[top]) * (i - st[len(st)-1] - 1) + res += tmp + } + } + st = append(st, i) + } + } + return res +} +func min(x, y int) int { + if x >= y { + return y + } + return x +} +``` + ### JavaScript: ```javascript From 9a068272ce8fd245d0c0ef93aee689afe4ea3cce Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Fri, 10 Jun 2022 22:46:32 +0800 Subject: [PATCH 042/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880129.?= =?UTF-8?q?=E6=B1=82=E6=A0=B9=E5=88=B0=E5=8F=B6=E5=AD=90=E8=8A=82=E7=82=B9?= =?UTF-8?q?=E6=95=B0=E5=AD=97=E4=B9=8B=E5=92=8C.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0129.求根到叶子节点数字之和.md | 33 +++++++++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/0129.求根到叶子节点数字之和.md b/problems/0129.求根到叶子节点数字之和.md index b271ca7d..a34e6921 100644 --- a/problems/0129.求根到叶子节点数字之和.md +++ b/problems/0129.求根到叶子节点数字之和.md @@ -289,7 +289,40 @@ var sumNumbers = function(root) { }; ``` +TypeScript： + +```typescript +function sumNumbers(root: TreeNode | null): number { + if (root === null) return 0; + let resTotal: number = 0; + const route: number[] = []; + route.push(root.val); + recur(root, route); + return resTotal; + function recur(node: TreeNode, route: number[]): void { + if (node.left === null && node.right === null) { + resTotal += listToSum(route); + return; + } + if (node.left !== null) { + route.push(node.left.val); + recur(node.left, route); + route.pop(); + }; + if (node.right !== null) { + route.push(node.right.val); + recur(node.right, route); + route.pop(); + }; + } + function listToSum(nums: number[]): number { + return Number(nums.join('')); + } +}; +``` + C: + ```c //sum记录总和 int sum; From 192beffb66d20fb51cdb61f11649e5cff6379bb8 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Fri, 10 Jun 2022 23:38:32 +0800 Subject: [PATCH 043/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=881382.?= =?UTF-8?q?=E5=B0=86=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E5=8F=98?= =?UTF-8?q?=E5=B9=B3=E8=A1=A1.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typesc?= =?UTF-8?q?ript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1382.将二叉搜索树变平衡.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/1382.将二叉搜索树变平衡.md b/problems/1382.将二叉搜索树变平衡.md index 57231ec4..d4d60ebd 100644 --- a/problems/1382.将二叉搜索树变平衡.md +++ b/problems/1382.将二叉搜索树变平衡.md @@ -148,6 +148,30 @@ var balanceBST = function(root) { }; ``` +TypeScript： + +```typescript +function balanceBST(root: TreeNode | null): TreeNode | null { + const inorderArr: number[] = []; + inorderTraverse(root, inorderArr); + return buildTree(inorderArr, 0, inorderArr.length - 1); +}; +function inorderTraverse(node: TreeNode | null, arr: number[]): void { + if (node === null) return; + inorderTraverse(node.left, arr); + arr.push(node.val); + inorderTraverse(node.right, arr); +} +function buildTree(arr: number[], left: number, right: number): TreeNode | null { + if (left > right) return null; + const mid = (left + right) >> 1; + const resNode: TreeNode = new TreeNode(arr[mid]); + resNode.left = buildTree(arr, left, mid - 1); + resNode.right = buildTree(arr, mid + 1, right); + return resNode; +} +``` + ----------------------- From 2386694b5b67e1ec17e42ce2c1f9df784eaeab24 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 11 Jun 2022 00:43:30 +0800 Subject: [PATCH 044/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880100.?= =?UTF-8?q?=E7=9B=B8=E5=90=8C=E7=9A=84=E6=A0=91.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0100.相同的树.md | 40 +++++++++++++++++++++++++++++++++++++++ 1 file changed, 40 insertions(+) diff --git a/problems/0100.相同的树.md b/problems/0100.相同的树.md index 5e805d01..d2431f39 100644 --- a/problems/0100.相同的树.md +++ b/problems/0100.相同的树.md @@ -240,6 +240,46 @@ Go： JavaScript： +TypeScript: + +> 递归法-先序遍历 + +```typescript +function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean { + if (p === null && q === null) return true; + if (p === null || q === null) return false; + if (p.val !== q.val) return false; + return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); +}; +``` + +> 迭代法-层序遍历 + +```typescript +function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean { + const queue1: (TreeNode | null)[] = [], + queue2: (TreeNode | null)[] = []; + queue1.push(p); + queue2.push(q); + while (queue1.length > 0 && queue2.length > 0) { + const node1 = queue1.shift(), + node2 = queue2.shift(); + if (node1 === null && node2 === null) continue; + if ( + (node1 === null || node2 === null) || + node1!.val !== node2!.val + ) return false; + queue1.push(node1!.left); + queue1.push(node1!.right); + queue2.push(node2!.left); + queue2.push(node2!.right); + } + return true; +}; +``` + + + -----------------------
From bd4f69d04b663dae39ab5a384722cd7d5aae3377 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 11 Jun 2022 16:35:06 +0800 Subject: [PATCH 045/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200053.=E6=9C=80?= =?UTF-8?q?=E5=A4=A7=E5=AD=90=E5=BA=8F=E5=92=8C=EF=BC=88=E5=8A=A8=E6=80=81?= =?UTF-8?q?=E8=A7=84=E5=88=92=EF=BC=89.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0053.最大子序和（动态规划）.md | 15 +++++++++++++++ 1 file changed, 15 insertions(+) diff --git a/problems/0053.最大子序和（动态规划）.md b/problems/0053.最大子序和（动态规划）.md index 4c883cb6..6712d0ac 100644 --- a/problems/0053.最大子序和（动态规划）.md +++ b/problems/0053.最大子序和（动态规划）.md @@ -186,7 +186,22 @@ const maxSubArray = nums => { }; ``` +Scala: +```scala +object Solution { + def maxSubArray(nums: Array[Int]): Int = { + var dp = new Array[Int](nums.length) + var result = nums(0) + dp(0) = nums(0) + for (i <- 1 until nums.length) { + dp(i) = math.max(nums(i), dp(i - 1) + nums(i)) + result = math.max(result, dp(i)) // 更新最大值 + } + result + } +} +``` -----------------------
From 9031523bbb334cb2190b17f12333d3a5f1edda53 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 11 Jun 2022 16:40:33 +0800 Subject: [PATCH 046/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200053.=E6=9C=80?= =?UTF-8?q?=E5=A4=A7=E5=AD=90=E5=BA=8F=E5=92=8C.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0053.最大子序和.md | 33 +++++++++++++++++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index 73cac244..4971133f 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -333,8 +333,41 @@ function maxSubArray(nums: number[]): number { }; ``` +### Scala +**贪心** +```scala +object Solution { + def maxSubArray(nums: Array[Int]): Int = { + var result = Int.MinValue + var count = 0 + for (i <- nums.indices) { + count += nums(i) // count累加 + if (count > result) result = count // 记录最大值 + if (count <= 0) count = 0 // 一旦count为负，则count归0 + } + result + } +} +``` + +**动态规划** + +```scala +object Solution { + def maxSubArray(nums: Array[Int]): Int = { + var dp = new Array[Int](nums.length) + var result = nums(0) + dp(0) = nums(0) + for (i <- 1 until nums.length) { + dp(i) = math.max(nums(i), dp(i - 1) + nums(i)) + result = math.max(result, dp(i)) // 更新最大值 + } + result + } +} +``` -----------------------
From ff233a8e3ccfb08e82ebaa662c93bf687b76f153 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 11 Jun 2022 16:53:35 +0800 Subject: [PATCH 047/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200122.=E4=B9=B0?= =?UTF-8?q?=E5=8D=96=E8=82=A1=E7=A5=A8=E7=9A=84=E6=9C=80=E4=BD=B3=E6=97=B6?= =?UTF-8?q?=E6=9C=BAII.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0122.买卖股票的最佳时机II.md | 35 +++++++++++++++++++++------ 1 file changed, 27 insertions(+), 8 deletions(-) diff --git a/problems/0122.买卖股票的最佳时机II.md b/problems/0122.买卖股票的最佳时机II.md index 1e7b77d8..1369ff5b 100644 --- a/problems/0122.买卖股票的最佳时机II.md +++ b/problems/0122.买卖股票的最佳时机II.md @@ -133,8 +133,9 @@ public: ## 其他语言版本 -Java: +### Java: +贪心: ```java // 贪心思路 class Solution { @@ -148,6 +149,7 @@ class Solution { } ``` +动态规划: ```java class Solution { // 动态规划 public int maxProfit(int[] prices) { @@ -169,8 +171,8 @@ class Solution { // 动态规划 } ``` -Python: - +### Python: +贪心: ```python class Solution: def maxProfit(self, prices: List[int]) -> int: @@ -180,7 +182,7 @@ class Solution: return result ``` -python动态规划 +动态规划: ```python class Solution: def maxProfit(self, prices: List[int]) -> int: @@ -194,7 +196,7 @@ class Solution: return dp[-1][1] ``` -Go: +### Go: ```golang //贪心算法 @@ -231,7 +233,7 @@ func maxProfit(prices []int) int { } ``` -Javascript: +### Javascript: 贪心 ```Javascript @@ -268,7 +270,7 @@ const maxProfit = (prices) => { }; ``` -TypeScript： +### TypeScript： ```typescript function maxProfit(prices: number[]): number { @@ -280,7 +282,7 @@ function maxProfit(prices: number[]): number { }; ``` -C: +### C: 贪心： ```c int maxProfit(int* prices, int pricesSize){ @@ -318,5 +320,22 @@ int maxProfit(int* prices, int pricesSize){ } ``` +### Scala + +贪心: +```scala +object Solution { + def maxProfit(prices: Array[Int]): Int = { + var result = 0 + for (i <- 1 until prices.length) { + if (prices(i) > prices(i - 1)) { + result += prices(i) - prices(i - 1) + } + } + result + } +} +``` + -----------------------
From b0664fcb81c475356ddf925b839df6e00c434e8a Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 11 Jun 2022 17:21:03 +0800 Subject: [PATCH 048/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200055.=E8=B7=B3?= =?UTF-8?q?=E8=B7=83=E6=B8=B8=E6=88=8F.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0055.跳跃游戏.md | 17 ++++++++++++++++- 1 file changed, 16 insertions(+), 1 deletion(-) diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index 17a3b4f4..345f8eba 100644 --- a/problems/0055.跳跃游戏.md +++ b/problems/0055.跳跃游戏.md @@ -193,7 +193,22 @@ function canJump(nums: number[]): boolean { }; ``` - +### Scala +```scala +object Solution { + def canJump(nums: Array[Int]): Boolean = { + var cover = 0 + if (nums.length == 1) return true // 如果只有一个元素，那么必定到达 + var i = 0 + while (i <= cover) { // i表示下标，当前只能够走cover步 + cover = math.max(i + nums(i), cover) + if (cover >= nums.length - 1) return true // 说明可以覆盖到终点，直接返回 + i += 1 + } + false // 如果上面没有返回就是跳不到 + } +} +``` ----------------------- From ad24b1fb1fa50e568a7357bbe772df59fc008e74 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 11 Jun 2022 18:29:48 +0800 Subject: [PATCH 049/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880116.?= =?UTF-8?q?=E5=A1=AB=E5=85=85=E6=AF=8F=E4=B8=AA=E8=8A=82=E7=82=B9=E7=9A=84?= =?UTF-8?q?=E4=B8=8B=E4=B8=80=E4=B8=AA=E5=8F=B3=E4=BE=A7=E8=8A=82=E7=82=B9?= =?UTF-8?q?=E6=8C=87=E9=92=88.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typesc?= =?UTF-8?q?ript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0116.填充每个节点的下一个右侧节点指针.md | 73 +++++++++++++++++++ 1 file changed, 73 insertions(+) diff --git a/problems/0116.填充每个节点的下一个右侧节点指针.md b/problems/0116.填充每个节点的下一个右侧节点指针.md index 2c443de5..1e5b2271 100644 --- a/problems/0116.填充每个节点的下一个右侧节点指针.md +++ b/problems/0116.填充每个节点的下一个右侧节点指针.md @@ -287,6 +287,79 @@ const connect = root => { }; ``` +## TypeScript + +（注：命名空间‘Node’与typescript中内置类型冲突，这里改成了‘NodePro’） + +> 递归法： + +```typescript +class NodePro { + val: number + left: NodePro | null + right: NodePro | null + next: NodePro | null + constructor(val?: number, left?: NodePro, right?: NodePro, next?: NodePro) { + this.val = (val === undefined ? 0 : val) + this.left = (left === undefined ? null : left) + this.right = (right === undefined ? null : right) + this.next = (next === undefined ? null : next) + } +} + +function connect(root: NodePro | null): NodePro | null { + if (root === null) return null; + root.next = null; + recur(root); + return root; +}; +function recur(node: NodePro): void { + if (node.left === null || node.right === null) return; + node.left.next = node.right; + node.right.next = node.next && node.next.left; + recur(node.left); + recur(node.right); +} +``` + +> 迭代法： + +```typescript +class NodePro { + val: number + left: NodePro | null + right: NodePro | null + next: NodePro | null + constructor(val?: number, left?: NodePro, right?: NodePro, next?: NodePro) { + this.val = (val === undefined ? 0 : val) + this.left = (left === undefined ? null : left) + this.right = (right === undefined ? null : right) + this.next = (next === undefined ? null : next) + } +} + +function connect(root: NodePro | null): NodePro | null { + if (root === null) return null; + const queue: NodePro[] = []; + queue.push(root); + while (queue.length > 0) { + for (let i = 0, length = queue.length; i < length; i++) { + const curNode: NodePro = queue.shift()!; + if (i === length - 1) { + curNode.next = null; + } else { + curNode.next = queue[0]; + } + if (curNode.left !== null) queue.push(curNode.left); + if (curNode.right !== null) queue.push(curNode.right); + } + } + return root; +}; +``` + + + -----------------------
From 74a422c53b2817271b8f50a61bdd8ccdf77a5e25 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 11 Jun 2022 20:03:12 +0800 Subject: [PATCH 050/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200045.=E8=B7=B3?= =?UTF-8?q?=E8=B7=83=E6=B8=B8=E6=88=8FII.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0045.跳跃游戏II.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0045.跳跃游戏II.md b/problems/0045.跳跃游戏II.md index 4e3ab24a..f2940361 100644 --- a/problems/0045.跳跃游戏II.md +++ b/problems/0045.跳跃游戏II.md @@ -279,7 +279,31 @@ function jump(nums: number[]): number { }; ``` +### Scala +```scala +object Solution { + def jump(nums: Array[Int]): Int = { + if (nums.length == 0) return 0 + var result = 0 // 记录走的最大步数 + var curDistance = 0 // 当前覆盖最远距离下标 + var nextDistance = 0 // 下一步覆盖最远距离下标 + for (i <- nums.indices) { + nextDistance = math.max(nums(i) + i, nextDistance) // 更新下一步覆盖最远距离下标 + if (i == curDistance) { + if (curDistance != nums.length - 1) { + result += 1 + curDistance = nextDistance + if (nextDistance >= nums.length - 1) return result + } else { + return result + } + } + } + result + } +} +``` From 67f74cdc098c7e7b7e012dec82d626731a59b051 Mon Sep 17 00:00:00 2001 From: tianzhou Date: Sat, 11 Jun 2022 23:08:31 +0800 Subject: [PATCH 051/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E5=AE=9E=E6=97=B6=20?= =?UTF-8?q?star=20history=20=E5=9B=BE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- README.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/README.md b/README.md index 4e2993d8..620942dc 100644 --- a/README.md +++ b/README.md @@ -523,6 +523,10 @@ [点此这里](https://github.com/youngyangyang04/leetcode-master/graphs/contributors)查看LeetCode-Master的所有贡献者。感谢他们补充了LeetCode-Master的其他语言版本，让更多的读者收益于此项目。 +# Star 趋势 + +[![Star History Chart](https://api.star-history.com/svg?repos=youngyangyang04/leetcode-master&type=Date)](https://star-history.com/#youngyangyang04/leetcode-master&Date) + # 关于作者 大家好，我是程序员Carl，哈工大师兄，《代码随想录》作者，先后在腾讯和百度从事后端技术研发，CSDN博客专家。对算法和C++后端技术有一定的见解，利用工作之余重新刷leetcode。 From 5d4046cacb2693a4e67fc311e2e83f73275d4909 Mon Sep 17 00:00:00 2001 From: guangyusong <15316444+guangyusong@users.noreply.github.com> Date: Sat, 11 Jun 2022 15:22:31 -0400 Subject: [PATCH 052/136] =?UTF-8?q?=E6=9B=B4=E6=96=B0server.md=E4=B8=ADUbu?= =?UTF-8?q?ntu=E7=9A=84=E6=8B=BC=E5=86=99=E9=94=99=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/qita/server.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/qita/server.md b/problems/qita/server.md index 16995d70..0748c104 100644 --- a/problems/qita/server.md +++ b/problems/qita/server.md @@ -105,7 +105,7 @@ https://github.com/youngyangyang04/fileHttpServer 如果你有一个服务器，那就是独立的一台电脑，你怎么霍霍就怎么霍霍，而且一年都不用关机的，可以一直跑你的任务，和你本地电脑也完全隔离。 -更方便的是，你目前系统假如是centos，想做一个实验需要在unbantu上，如果是云服务器，更换系统就是在 后台点一下，一键重装，云厂商基本都是支持所有系统一件安装的。 +更方便的是，你目前系统假如是CentOS，想做一个实验需要在Ubuntu上，如果是云服务器，更换系统就是在 后台点一下，一键重装，云厂商基本都是支持所有系统一件安装的。 我们平时自己玩linux经常是配各种环境，然后这个linux就被自己玩坏了（一般都是毫无节制使用root权限导致的），总之就是环境配不起来了，基本就要重装了。 From 882c19c3e263161a18d4349f53ee2226a09a3c6b Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sun, 12 Jun 2022 10:17:49 +0800 Subject: [PATCH 053/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201005.K=E6=AC=A1?= =?UTF-8?q?=E5=8F=96=E5=8F=8D=E5=90=8E=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84?= =?UTF-8?q?=E6=95=B0=E7=BB=84=E5=92=8C.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1005.K次取反后最大化的数组和.md | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md index 202534da..71fc628f 100644 --- a/problems/1005.K次取反后最大化的数组和.md +++ b/problems/1005.K次取反后最大化的数组和.md @@ -289,6 +289,28 @@ function largestSumAfterKNegations(nums: number[], k: number): number { }; ``` +### Scala + +```scala +object Solution { + def largestSumAfterKNegations(nums: Array[Int], k: Int): Int = { + var num = nums.sortWith(math.abs(_) > math.abs(_)) + + var kk = k // 因为k是不可变量，所以要赋值给一个可变量 + for (i <- num.indices) { + if (num(i) < 0 && kk > 0) { + num(i) *= -1 // 取反 + kk -= 1 + } + } + + // kk对2取余，结果为0则为偶数不需要取反，结果为1为奇数，只需要对最后的数字进行反转就可以 + if (kk % 2 == 1) num(num.size - 1) *= -1 + + num.sum // 最后返回数字的和 + } +} +``` From dd20ca032fab47e0e05f9d32c05f528872dcf83a Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 12 Jun 2022 12:17:26 +0800 Subject: [PATCH 054/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880052.N?= =?UTF-8?q?=E7=9A=87=E5=90=8EII.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0type?= =?UTF-8?q?script=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0052.N皇后II.md | 54 ++++++++++++++++++++++++++++++++++++++++ 1 file changed, 54 insertions(+) diff --git a/problems/0052.N皇后II.md b/problems/0052.N皇后II.md index 67e439ca..608aeda1 100644 --- a/problems/0052.N皇后II.md +++ b/problems/0052.N皇后II.md @@ -144,7 +144,61 @@ var totalNQueens = function(n) { }; ``` +TypeScript： + +```typescript +// 0-该格为空，1-该格有皇后 +type GridStatus = 0 | 1; +function totalNQueens(n: number): number { + let resCount: number = 0; + const chess: GridStatus[][] = new Array(n).fill(0) + .map(_ => new Array(n).fill(0)); + backTracking(chess, n, 0); + return resCount; + function backTracking(chess: GridStatus[][], n: number, startRowIndex: number): void { + if (startRowIndex === n) { + resCount++; + return; + } + for (let j = 0; j < n; j++) { + if (checkValid(chess, startRowIndex, j, n) === true) { + chess[startRowIndex][j] = 1; + backTracking(chess, n, startRowIndex + 1); + chess[startRowIndex][j] = 0; + } + } + } +}; +function checkValid(chess: GridStatus[][], i: number, j: number, n: number): boolean { + // 向上纵向检查 + let tempI: number = i - 1, + tempJ: number = j; + while (tempI >= 0) { + if (chess[tempI][tempJ] === 1) return false; + tempI--; + } + // 斜向左上检查 + tempI = i - 1; + tempJ = j - 1; + while (tempI >= 0 && tempJ >= 0) { + if (chess[tempI][tempJ] === 1) return false; + tempI--; + tempJ--; + } + // 斜向右上检查 + tempI = i - 1; + tempJ = j + 1; + while (tempI >= 0 && tempJ < n) { + if (chess[tempI][tempJ] === 1) return false; + tempI--; + tempJ++; + } + return true; +} +``` + C + ```c //path[i]为在i行，path[i]列上存在皇后 int *path; From 1da68a332c35b178d62edf946fd7a4da4ac98d3c Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sun, 12 Jun 2022 13:46:58 +0800 Subject: [PATCH 055/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200134.=E5=8A=A0?= =?UTF-8?q?=E6=B2=B9=E7=AB=99.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0134.加油站.md | 68 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 68 insertions(+) diff --git a/problems/0134.加油站.md b/problems/0134.加油站.md index a88f677d..541be293 100644 --- a/problems/0134.加油站.md +++ b/problems/0134.加油站.md @@ -471,5 +471,73 @@ int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){ } ``` +### Scala + +暴力解法: + +```scala +object Solution { + def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = { + for (i <- cost.indices) { + var rest = gas(i) - cost(i) + var index = (i + 1) % cost.length // index为i的下一个节点 + while (rest > 0 && i != index) { + rest += (gas(index) - cost(index)) + index = (index + 1) % cost.length + } + if (rest >= 0 && index == i) return i + } + -1 + } +} +``` + +贪心算法，方法一: + +```scala +object Solution { + def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = { + var curSum = 0 + var min = Int.MaxValue + for (i <- gas.indices) { + var rest = gas(i) - cost(i) + curSum += rest + min = math.min(min, curSum) + } + if (curSum < 0) return -1 // 情况1: gas的总和小于cost的总和，不可能到达终点 + if (min >= 0) return 0 // 情况2: 最小值>=0，从0号出发可以直接到达 + // 情况3: min为负值，从后向前看，能把min填平的节点就是出发节点 + for (i <- gas.length - 1 to 0 by -1) { + var rest = gas(i) - cost(i) + min += rest + if (min >= 0) return i + } + -1 + } +} +``` + +贪心算法，方法二: + +```scala +object Solution { + def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = { + var curSum = 0 + var totalSum = 0 + var start = 0 + for (i <- gas.indices) { + curSum += (gas(i) - cost(i)) + totalSum += (gas(i) - cost(i)) + if (curSum < 0) { + start = i + 1 // 起始位置更新 + curSum = 0 // curSum从0开始 + } + } + if (totalSum < 0) return -1 // 说明怎么走不可能跑一圈 + start + } +} +``` + -----------------------
From 998785bcab1346d652b8b3db9ccd166054194a83 Mon Sep 17 00:00:00 2001 From: Vincent Date: Sun, 12 Jun 2022 16:59:08 -0700 Subject: [PATCH 056/136] =?UTF-8?q?Update=200034.=E5=9C=A8=E6=8E=92?= =?UTF-8?q?=E5=BA=8F=E6=95=B0=E7=BB=84=E4=B8=AD=E6=9F=A5=E6=89=BE=E5=85=83?= =?UTF-8?q?=E7=B4=A0=E7=9A=84=E7=AC=AC=E4=B8=80=E4=B8=AA=E5=92=8C=E6=9C=80?= =?UTF-8?q?=E5=90=8E=E4=B8=80=E4=B8=AA=E4=BD=8D=E7=BD=AE.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加力扣原题链接 --- problems/0034.在排序数组中查找元素的第一个和最后一个位置.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md index 260462c2..b6e82262 100644 --- a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md +++ b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md @@ -7,6 +7,8 @@ # 34. 在排序数组中查找元素的第一个和最后一个位置 +[力扣链接](https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/) + 给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。 如果数组中不存在目标值 target，返回 [-1, -1]。 From 0a3a2d4ae2213cad03d8f339f09f4e9de7420998 Mon Sep 17 00:00:00 2001 From: Rinko Taketsuki <33001553+RinkoTaketsuki@users.noreply.github.com> Date: Mon, 13 Jun 2022 09:27:00 +0800 Subject: [PATCH 057/136] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index a51c68ee..1d8a0e64 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -49,7 +49,7 @@ 数组长度为：len。 -如果len % (len - (next[len - 1] + 1)) == 0 ，则说明 (数组长度-最长相等前后缀的长度) 正好可以被 数组的长度整除，说明有该字符串有重复的子字符串。 +如果len % (len - (next[len - 1] + 1)) == 0 ，则说明数组的长度正好可以被 (数组长度-最长相等前后缀的长度) 整除 ，说明该字符串有重复的子字符串。 **数组长度减去最长相同前后缀的长度相当于是第一个周期的长度，也就是一个周期的长度，如果这个周期可以被整除，就说明整个数组就是这个周期的循环。** From 32018ff79afa8fb65cdf69445c9d93a34e648d54 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Mon, 13 Jun 2022 20:53:11 +0800 Subject: [PATCH 058/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200135.=E5=88=86?= =?UTF-8?q?=E5=8F=91=E7=B3=96=E6=9E=9C.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0135.分发糖果.md | 25 +++++++++++++++++++++++++ 1 file changed, 25 insertions(+) diff --git a/problems/0135.分发糖果.md b/problems/0135.分发糖果.md index 3456a04c..a805d0d4 100644 --- a/problems/0135.分发糖果.md +++ b/problems/0135.分发糖果.md @@ -324,6 +324,31 @@ function candy(ratings: number[]): number { }; ``` +### Scala + +```scala +object Solution { + def candy(ratings: Array[Int]): Int = { + var candyVec = new Array[Int](ratings.length) + for (i <- candyVec.indices) candyVec(i) = 1 + // 从前向后 + for (i <- 1 until candyVec.length) { + if (ratings(i) > ratings(i - 1)) { + candyVec(i) = candyVec(i - 1) + 1 + } + } + + // 从后向前 + for (i <- (candyVec.length - 2) to 0 by -1) { + if (ratings(i) > ratings(i + 1)) { + candyVec(i) = math.max(candyVec(i), candyVec(i + 1) + 1) + } + } + + candyVec.sum // 求和 + } +} +``` ----------------------- From 6867c9c5bfd379a2f6e8afb09026cb0e759bc8dc Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Mon, 13 Jun 2022 21:21:22 +0800 Subject: [PATCH 059/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200860.=E6=9F=A0?= =?UTF-8?q?=E6=AA=AC=E6=B0=B4=E6=89=BE=E9=9B=B6.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0860.柠檬水找零.md | 31 +++++++++++++++++++++++++++++++ 1 file changed, 31 insertions(+) diff --git a/problems/0860.柠檬水找零.md b/problems/0860.柠檬水找零.md index aa09e1c6..4a676b43 100644 --- a/problems/0860.柠檬水找零.md +++ b/problems/0860.柠檬水找零.md @@ -328,6 +328,37 @@ function lemonadeChange(bills: number[]): boolean { ``` +### Scala + +```scala +object Solution { + def lemonadeChange(bills: Array[Int]): Boolean = { + var fiveNum = 0 + var tenNum = 0 + + for (i <- bills) { + if (i == 5) fiveNum += 1 + if (i == 10) { + if (fiveNum <= 0) return false + tenNum += 1 + fiveNum -= 1 + } + if (i == 20) { + if (fiveNum > 0 && tenNum > 0) { + tenNum -= 1 + fiveNum -= 1 + } else if (fiveNum >= 3) { + fiveNum -= 3 + } else { + return false + } + } + } + true + } +} +``` + -----------------------
From 17208d89290c4bc10256414393671022b6b96ccf Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Mon, 13 Jun 2022 22:12:49 +0800 Subject: [PATCH 060/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200406.=E6=A0=B9?= =?UTF-8?q?=E6=8D=AE=E8=BA=AB=E9=AB=98=E9=87=8D=E5=BB=BA=E9=98=9F=E5=88=97?= =?UTF-8?q?.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0406.根据身高重建队列.md | 19 +++++++++++++++++++ 1 file changed, 19 insertions(+) diff --git a/problems/0406.根据身高重建队列.md b/problems/0406.根据身高重建队列.md index 641086a9..516df7d7 100644 --- a/problems/0406.根据身高重建队列.md +++ b/problems/0406.根据身高重建队列.md @@ -354,8 +354,27 @@ function reconstructQueue(people: number[][]): number[][] { }; ``` +### Scala +```scala +object Solution { + import scala.collection.mutable + def reconstructQueue(people: Array[Array[Int]]): Array[Array[Int]] = { + val person = people.sortWith((a, b) => { + if (a(0) == b(0)) a(1) < b(1) + else a(0) > b(0) + }) + var que = mutable.ArrayBuffer[Array[Int]]() + + for (per <- person) { + que.insert(per(1), per) + } + + que.toArray + } +} +``` ----------------------- From e391bf18f7f0c607a918413d26866a9f376ced1e Mon Sep 17 00:00:00 2001 From: van_fantasy <46948123+sexyxlyWAol@users.noreply.github.com> Date: Mon, 13 Jun 2022 22:44:49 +0800 Subject: [PATCH 061/136] =?UTF-8?q?Update=201221.=E5=88=86=E5=89=B2?= =?UTF-8?q?=E5=B9=B3=E8=A1=A1=E5=AD=97=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit added python and go solution to 1221 --- problems/1221.分割平衡字符串.md | 27 +++++++++++++++++++++++++++ 1 file changed, 27 insertions(+) diff --git a/problems/1221.分割平衡字符串.md b/problems/1221.分割平衡字符串.md index 1a9b34a2..e18c6358 100644 --- a/problems/1221.分割平衡字符串.md +++ b/problems/1221.分割平衡字符串.md @@ -108,11 +108,38 @@ class Solution { ### Python ```python +class Solution: + def balancedStringSplit(self, s: str) -> int: + diff = 0 #右左差值 + ans = 0 + for c in s: + if c == "L": + diff -= 1 + else: + diff += 1 + if tilt == 0: + ans += 1 + return ans ``` ### Go ```go +func balancedStringSplit(s string) int { + diff := 0 // 右左差值 + ans := 0 + for _, c := range s { + if c == 'L' { + diff-- + }else { + diff++ + } + if diff == 0 { + ans++ + } + } + return ans +} ``` ### JavaScript From 8f52696f2206f2361917aad3e1cc7f87698c5cc0 Mon Sep 17 00:00:00 2001 From: Rinko Taketsuki <33001553+RinkoTaketsuki@users.noreply.github.com> Date: Tue, 14 Jun 2022 09:24:58 +0800 Subject: [PATCH 062/136] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 1d8a0e64..ba2fa160 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -63,7 +63,7 @@ next[len - 1] = 7，next[len - 1] + 1 = 8，8就是此时字符串asdfasdfasdf的最长相同前后缀的长度。 -(len - (next[len - 1] + 1)) 也就是： 12(字符串的长度) - 8(最长公共前后缀的长度) = 4， 4正好可以被 12(字符串的长度) 整除，所以说明有重复的子字符串（asdf）。 +(len - (next[len - 1] + 1)) 也就是： 12(字符串的长度) - 8(最长公共前后缀的长度) = 4， 12 正好可以被 4 整除，所以说明有重复的子字符串（asdf）。 C++代码如下：（这里使用了前缀表统一减一的实现方式） From e8a8db47cfc42380b0979e8efcd7f9908216b2a6 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Tue, 14 Jun 2022 20:03:34 +0800 Subject: [PATCH 063/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200452.=E7=94=A8?= =?UTF-8?q?=E6=9C=80=E5=B0=91=E6=95=B0=E9=87=8F=E7=9A=84=E7=AE=AD=E5=BC=95?= =?UTF-8?q?=E7=88=86=E6=B0=94=E7=90=83.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0452.用最少数量的箭引爆气球.md | 25 +++++++++++++++++++++++++ 1 file changed, 25 insertions(+) diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md index d4bbe961..e07aa6e6 100644 --- a/problems/0452.用最少数量的箭引爆气球.md +++ b/problems/0452.用最少数量的箭引爆气球.md @@ -288,5 +288,30 @@ impl Solution { } } ``` + +### Scala + +```scala +object Solution { + def findMinArrowShots(points: Array[Array[Int]]): Int = { + if (points.length == 0) return 0 + // 排序 + var point = points.sortWith((a, b) => { + a(0) < b(0) + }) + + var result = 1 // points不为空就至少需要一只箭 + for (i <- 1 until point.length) { + if (point(i)(0) > point(i - 1)(1)) { + result += 1 + } else { + point(i)(1) = math.min(point(i - 1)(1), point(i)(1)) + } + } + result // 返回结果 + } +} +``` + -----------------------
From ece2c3efb6e492529c4c8b9b4be1bdf7fbb1b8b6 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Tue, 14 Jun 2022 20:45:13 +0800 Subject: [PATCH 064/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200435.=E6=97=A0?= =?UTF-8?q?=E9=87=8D=E5=8F=A0=E5=8C=BA=E9=97=B4.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0435.无重叠区间.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/problems/0435.无重叠区间.md b/problems/0435.无重叠区间.md index 66aa1244..6313bc44 100644 --- a/problems/0435.无重叠区间.md +++ b/problems/0435.无重叠区间.md @@ -352,7 +352,27 @@ function eraseOverlapIntervals(intervals: number[][]): number { }; ``` +### Scala +```scala +object Solution { + def eraseOverlapIntervals(intervals: Array[Array[Int]]): Int = { + var result = 0 + var interval = intervals.sortWith((a, b) => { + a(1) < b(1) + }) + var edge = Int.MinValue + for (i <- 0 until interval.length) { + if (edge <= interval(i)(0)) { + edge = interval(i)(1) + } else { + result += 1 + } + } + result + } +} +``` From 94350c0c99107f5343a589d36382e52a8a3c0059 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Tue, 14 Jun 2022 22:26:56 +0800 Subject: [PATCH 065/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200763.=E5=88=92?= =?UTF-8?q?=E5=88=86=E5=AD=97=E6=AF=8D=E5=8C=BA=E9=97=B4.md=20Scala?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0763.划分字母区间.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0763.划分字母区间.md b/problems/0763.划分字母区间.md index 2f4d1b48..f7b16f5c 100644 --- a/problems/0763.划分字母区间.md +++ b/problems/0763.划分字母区间.md @@ -317,7 +317,31 @@ function partitionLabels(s: string): number[] { }; ``` +### Scala +```scala +object Solution { + import scala.collection.mutable + def partitionLabels(s: String): List[Int] = { + var hash = new Array[Int](26) + for (i <- s.indices) { + hash(s(i) - 'a') = i + } + + var res = mutable.ListBuffer[Int]() + var (left, right) = (0, 0) + for (i <- s.indices) { + right = math.max(hash(s(i) - 'a'), right) + if (i == right) { + res.append(right - left + 1) + left = i + 1 + } + } + + res.toList + } +} +``` ----------------------- From eddfde7c1cb4c7db4c11f7d6c76b8a732a65d313 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 15 Jun 2022 22:25:35 +0800 Subject: [PATCH 066/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200056.=E5=90=88?= =?UTF-8?q?=E5=B9=B6=E5=8C=BA=E9=97=B4.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0056.合并区间.md | 30 ++++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index e444a221..38e3472c 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -286,7 +286,37 @@ function merge(intervals: number[][]): number[][] { }; ``` +### Scala +```scala +object Solution { + import scala.collection.mutable + def merge(intervals: Array[Array[Int]]): Array[Array[Int]] = { + var res = mutable.ArrayBuffer[Array[Int]]() + + // 排序 + var interval = intervals.sortWith((a, b) => { + a(0) < b(0) + }) + + var left = interval(0)(0) + var right = interval(0)(1) + + for (i <- 1 until interval.length) { + if (interval(i)(0) <= right) { + left = math.min(left, interval(i)(0)) + right = math.max(right, interval(i)(1)) + } else { + res.append(Array[Int](left, right)) + left = interval(i)(0) + right = interval(i)(1) + } + } + res.append(Array[Int](left, right)) + res.toArray // 返回res的Array形式 + } +} +``` -----------------------
From 23c26135135f39757932a158d5c4b9f00f4bfe3b Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 15 Jun 2022 23:15:29 +0800 Subject: [PATCH 067/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200738.=E5=8D=95?= =?UTF-8?q?=E8=B0=83=E9=80=92=E5=A2=9E=E7=9A=84=E6=95=B0=E5=AD=97.md=20Sca?= =?UTF-8?q?la=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0738.单调递增的数字.md | 30 ++++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0738.单调递增的数字.md b/problems/0738.单调递增的数字.md index 4e4079a7..2911e1cc 100644 --- a/problems/0738.单调递增的数字.md +++ b/problems/0738.单调递增的数字.md @@ -246,7 +246,37 @@ function monotoneIncreasingDigits(n: number): number { ``` +### Scala +直接转换为了整数数组: +```scala +object Solution { + import scala.collection.mutable + def monotoneIncreasingDigits(n: Int): Int = { + var digits = mutable.ArrayBuffer[Int]() + // 提取每位数字 + var temp = n // 因为 参数n 是不可变量所以需要赋值给一个可变量 + while (temp != 0) { + digits.append(temp % 10) + temp = temp / 10 + } + // 贪心 + var flag = -1 + for (i <- 0 until (digits.length - 1) if digits(i) < digits(i + 1)) { + flag = i + digits(i + 1) -= 1 + } + for (i <- 0 to flag) digits(i) = 9 + + // 拼接 + var res = 0 + for (i <- 0 until digits.length) { + res += digits(i) * math.pow(10, i).toInt + } + res + } +} +``` -----------------------
From cff9cf34b0725b43598c82faccda24d83e0a06ea Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=96=9D=E9=86=89=E7=9A=84=E7=8E=A9=E5=85=B7=E7=86=8A=5F?= =?UTF-8?q?=E7=8E=8B=E5=9D=87=E7=A5=A5?= <1033076925@qq.com> Date: Thu, 16 Jun 2022 13:16:02 +0800 Subject: [PATCH 068/136] =?UTF-8?q?=E4=BF=AE=E6=AD=A30035.=E6=90=9C?= =?UTF-8?q?=E7=B4=A2=E6=8F=92=E5=85=A5=E4=BD=8D=E7=BD=AE=E9=94=99=E5=88=AB?= =?UTF-8?q?=E5=AD=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0035.搜索插入位置.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md index 5cf44ded..aef091af 100644 --- a/problems/0035.搜索插入位置.md +++ b/problems/0035.搜索插入位置.md @@ -142,7 +142,7 @@ public: ``` * 时间复杂度：O(log n) -* 时间复杂度：O(1) +* 空间复杂度：O(1) 效率如下： ![35_搜索插入位置2](https://img-blog.csdnimg.cn/2020121623272877.png) From a031937e874267a3a4dcbaf42eba272b244fc0fc Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Thu, 16 Jun 2022 22:32:13 +0800 Subject: [PATCH 069/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200714.=E4=B9=B0?= =?UTF-8?q?=E5=8D=96=E8=82=A1=E7=A5=A8=E7=9A=84=E6=9C=80=E4=BD=B3=E6=97=B6?= =?UTF-8?q?=E6=9C=BA=E5=90=AB=E6=89=8B=E7=BB=AD=E8=B4=B9.md=20Scala?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0714.买卖股票的最佳时机含手续费.md | 30 +++++++++++++++++---- 1 file changed, 25 insertions(+), 5 deletions(-) diff --git a/problems/0714.买卖股票的最佳时机含手续费.md b/problems/0714.买卖股票的最佳时机含手续费.md index b27631c6..4bc21a70 100644 --- a/problems/0714.买卖股票的最佳时机含手续费.md +++ b/problems/0714.买卖股票的最佳时机含手续费.md @@ -153,7 +153,7 @@ public: ## 其他语言版本 -Java： +### Java ```java // 贪心思路 class Solution { @@ -198,7 +198,7 @@ class Solution { // 动态规划 -Python： +### Python ```python class Solution: # 贪心思路 @@ -216,7 +216,7 @@ class Solution: # 贪心思路 return result ``` -Go： +### Go ```golang func maxProfit(prices []int, fee int) int { var minBuy int = prices[0] //第一天买入 @@ -241,7 +241,7 @@ func maxProfit(prices []int, fee int) int { return res } ``` -Javascript: +### Javascript ```Javascript // 贪心思路 var maxProfit = function(prices, fee) { @@ -293,7 +293,7 @@ var maxProfit = function(prices, fee) { }; ``` -TypeScript： +### TypeScript > 贪心 @@ -335,8 +335,28 @@ function maxProfit(prices: number[], fee: number): number { }; ``` +### Scala +贪心思路: +```scala +object Solution { + def maxProfit(prices: Array[Int], fee: Int): Int = { + var result = 0 + var minPrice = prices(0) + for (i <- 1 until prices.length) { + if (prices(i) < minPrice) { + minPrice = prices(i) // 比当前最小值还小 + } + if (prices(i) > minPrice + fee) { + result += prices(i) - minPrice - fee + minPrice = prices(i) - fee + } + } + result + } +} +``` -----------------------
From 0b9737d7541870e8f68d728fd0a0197d281b4c1d Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Thu, 16 Jun 2022 21:24:24 -0500 Subject: [PATCH 070/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A00028.=E5=AE=9E?= =?UTF-8?q?=E7=8E=B0strStr=20python=E7=89=88=E6=9C=AC=E6=9A=B4=E5=8A=9B?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 16 +++++++++++++++- 1 file changed, 15 insertions(+), 1 deletion(-) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index 1cdd5292..00997907 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -685,7 +685,21 @@ class Solution { ``` Python3： - +```python +//暴力解法： +class Solution(object): + def strStr(self, haystack, needle): + """ + :type haystack: str + :type needle: str + :rtype: int + """ + m,n=len(haystack),len(needle) + for i in range(m): + if haystack[i:i+n]==needle: + return i + return -1 +``` ```python // 方法一 class Solution: From da559f0ef013bb0503919bda084b244975a73ed6 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 18 Jun 2022 19:17:49 +0800 Subject: [PATCH 071/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200968.=E7=9B=91?= =?UTF-8?q?=E6=8E=A7=E4=BA=8C=E5=8F=89=E6=A0=91.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0968.监控二叉树.md | 35 +++++++++++++++++++++++++++++++++++ 1 file changed, 35 insertions(+) diff --git a/problems/0968.监控二叉树.md b/problems/0968.监控二叉树.md index 9a510a1b..6c957eb2 100644 --- a/problems/0968.监控二叉树.md +++ b/problems/0968.监控二叉树.md @@ -544,5 +544,40 @@ int minCameraCover(struct TreeNode* root){ } ``` +### Scala + +```scala +object Solution { + def minCameraCover(root: TreeNode): Int = { + var result = 0 + def traversal(cur: TreeNode): Int = { + // 空节点，该节点有覆盖 + if (cur == null) return 2 + var left = traversal(cur.left) + var right = traversal(cur.right) + // 情况1，左右节点都有覆盖 + if (left == 2 && right == 2) { + return 0 + } + // 情况2 + if (left == 0 || right == 0) { + result += 1 + return 1 + } + // 情况3 + if (left == 1 || right == 1) { + return 2 + } + -1 + } + + if (traversal(root) == 0) { + result += 1 + } + result + } +} +``` + -----------------------
From cf0affde0826eedf6328cf383d89d70028bf43a6 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 18 Jun 2022 19:30:51 +0800 Subject: [PATCH 072/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200509.=E6=96=90?= =?UTF-8?q?=E6=B3=A2=E9=82=A3=E5=A5=91=E6=95=B0.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0509.斐波那契数.md | 30 +++++++++++++++++++++++++++++- 1 file changed, 29 insertions(+), 1 deletion(-) diff --git a/problems/0509.斐波那契数.md b/problems/0509.斐波那契数.md index 1d17784d..d8e4e1d7 100644 --- a/problems/0509.斐波那契数.md +++ b/problems/0509.斐波那契数.md @@ -245,7 +245,7 @@ var fib = function(n) { }; ``` -TypeScript +### TypeScript ```typescript function fib(n: number): number { @@ -324,5 +324,33 @@ pub fn fib(n: i32) -> i32 { return fib(n - 1) + fib(n - 2); } ``` + +### Scala + +动态规划: +```scala +object Solution { + def fib(n: Int): Int = { + if (n <= 1) return n + var dp = new Array[Int](n + 1) + dp(1) = 1 + for (i <- 2 to n) { + dp(i) = dp(i - 1) + dp(i - 2) + } + dp(n) + } +} +``` + +递归: +```scala +object Solution { + def fib(n: Int): Int = { + if (n <= 1) return n + fib(n - 1) + fib(n - 2) + } +} +``` + -----------------------
From 5985aae83ce72f2d25b4976a836c1a5fd39ac1ef Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 18 Jun 2022 19:37:35 +0800 Subject: [PATCH 073/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200070.=E7=88=AC?= =?UTF-8?q?=E6=A5=BC=E6=A2=AF.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0070.爬楼梯.md | 32 ++++++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/problems/0070.爬楼梯.md b/problems/0070.爬楼梯.md index 34d41441..097466b0 100644 --- a/problems/0070.爬楼梯.md +++ b/problems/0070.爬楼梯.md @@ -401,6 +401,38 @@ int climbStairs(int n){ } ``` +### Scala + +```scala +object Solution { + def climbStairs(n: Int): Int = { + if (n <= 2) return n + var dp = new Array[Int](n + 1) + dp(1) = 1 + dp(2) = 2 + for (i <- 3 to n) { + dp(i) = dp(i - 1) + dp(i - 2) + } + dp(n) + } +} +``` + +优化空间复杂度: +```scala +object Solution { + def climbStairs(n: Int): Int = { + if (n <= 2) return n + var (a, b) = (1, 2) + for (i <- 3 to n) { + var tmp = a + b + a = b + b = tmp + } + b // 最终返回b + } +} +``` -----------------------
From 6e92cd2417dea438d7671140fd73298a4b598c93 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 18 Jun 2022 20:12:19 +0800 Subject: [PATCH 074/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200746.=E4=BD=BF?= =?UTF-8?q?=E7=94=A8=E6=9C=80=E5=B0=8F=E8=8A=B1=E8=B4=B9=E7=88=AC=E6=A5=BC?= =?UTF-8?q?=E6=A2=AF.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0746.使用最小花费爬楼梯.md | 30 +++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0746.使用最小花费爬楼梯.md b/problems/0746.使用最小花费爬楼梯.md index 5931fc8a..abaeb980 100644 --- a/problems/0746.使用最小花费爬楼梯.md +++ b/problems/0746.使用最小花费爬楼梯.md @@ -305,5 +305,35 @@ int minCostClimbingStairs(int* cost, int costSize){ return dp[i-1] < dp[i-2] ? dp[i-1] : dp[i-2]; } ``` + +### Scala + +```scala +object Solution { + def minCostClimbingStairs(cost: Array[Int]): Int = { + var dp = new Array[Int](cost.length) + dp(0) = cost(0) + dp(1) = cost(1) + for (i <- 2 until cost.length) { + dp(i) = math.min(dp(i - 1), dp(i - 2)) + cost(i) + } + math.min(dp(cost.length - 1), dp(cost.length - 2)) + } +} +``` + +第二种思路: dp[i] 表示爬到第i-1层所需的最小花费，状态转移方程为: dp[i] = min(dp[i-1]+cost[i-1],dp[i-2]+cost[i-2]) +```scala +object Solution { + def minCostClimbingStairs(cost: Array[Int]): Int = { + var dp = new Array[Int](cost.length + 1) + for (i <- 2 until cost.length + 1) { + dp(i) = math.min(dp(i - 1) + cost(i - 1), dp(i - 2) + cost(i - 2)) + } + dp(cost.length) + } +} +``` + -----------------------
From 8e27191aec0f9c377fe79b61a1b6deffa2af5fe4 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 18 Jun 2022 20:29:53 +0800 Subject: [PATCH 075/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200062.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E8=B7=AF=E5=BE=84.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0062.不同路径.md | 16 ++++++++++++++++ 1 file changed, 16 insertions(+) diff --git a/problems/0062.不同路径.md b/problems/0062.不同路径.md index f59b7be8..cccda7f1 100644 --- a/problems/0062.不同路径.md +++ b/problems/0062.不同路径.md @@ -412,5 +412,21 @@ int uniquePaths(int m, int n){ } ``` +### Scala + +```scala +object Solution { + def uniquePaths(m: Int, n: Int): Int = { + var dp = Array.ofDim[Int](m, n) + for (i <- 0 until m) dp(i)(0) = 1 + for (j <- 1 until n) dp(0)(j) = 1 + for (i <- 1 until m; j <- 1 until n) { + dp(i)(j) = dp(i - 1)(j) + dp(i)(j - 1) + } + dp(m - 1)(n - 1) + } +} +``` + -----------------------
From a9de01b83c02ab57181b52df37f4252b5709772f Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 18 Jun 2022 20:49:46 +0800 Subject: [PATCH 076/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200063.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E8=B7=AF=E5=BE=84II.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0063.不同路径II.md | 32 ++++++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/problems/0063.不同路径II.md b/problems/0063.不同路径II.md index 59c60156..88fce505 100644 --- a/problems/0063.不同路径II.md +++ b/problems/0063.不同路径II.md @@ -440,5 +440,37 @@ int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obst } ``` +### Scala + +```scala +object Solution { + import scala.util.control.Breaks._ + def uniquePathsWithObstacles(obstacleGrid: Array[Array[Int]]): Int = { + var (m, n) = (obstacleGrid.length, obstacleGrid(0).length) + var dp = Array.ofDim[Int](m, n) + + // 比如break、continue这些流程控制需要使用breakable + breakable( + for (i <- 0 until m) { + if (obstacleGrid(i)(0) != 1) dp(i)(0) = 1 + else break() + } + ) + breakable( + for (j <- 0 until n) { + if (obstacleGrid(0)(j) != 1) dp(0)(j) = 1 + else break() + } + ) + + for (i <- 1 until m; j <- 1 until n; if obstacleGrid(i)(j) != 1) { + dp(i)(j) = dp(i - 1)(j) + dp(i)(j - 1) + } + + dp(m - 1)(n - 1) + } +} +``` + -----------------------
From 568e36bae8ba2f2812a47a2d7548b9ebd20041b5 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 18 Jun 2022 21:45:03 +0800 Subject: [PATCH 077/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200343.=E6=95=B4?= =?UTF-8?q?=E6=95=B0=E6=8B=86=E5=88=86.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0343.整数拆分.md | 17 +++++++++++++++++ 1 file changed, 17 insertions(+) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index 279f1d71..9166f2cb 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -335,5 +335,22 @@ int integerBreak(int n){ } ``` +### Scala + +```scala +object Solution { + def integerBreak(n: Int): Int = { + var dp = new Array[Int](n + 1) + dp(2) = 1 + for (i <- 3 to n) { + for (j <- 1 until i - 1) { + dp(i) = math.max(dp(i), math.max(j * (i - j), j * dp(i - j))) + } + } + dp(n) + } +} +``` + -----------------------
From 374190e2232ec9daeffe20660d1020199a17c10b Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sat, 18 Jun 2022 22:25:08 +0800 Subject: [PATCH 078/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200096.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E7=9A=84=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91?= =?UTF-8?q?.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0096.不同的二叉搜索树.md | 19 ++++++++++++++++++- 1 file changed, 18 insertions(+), 1 deletion(-) diff --git a/problems/0096.不同的二叉搜索树.md b/problems/0096.不同的二叉搜索树.md index 25561b50..a33421ae 100644 --- a/problems/0096.不同的二叉搜索树.md +++ b/problems/0096.不同的二叉搜索树.md @@ -227,7 +227,7 @@ const numTrees =(n) => { }; ``` -TypeScript +### TypeScript ```typescript function numTrees(n: number): number { @@ -282,5 +282,22 @@ int numTrees(int n){ } ``` +### Scala + +```scala +object Solution { + def numTrees(n: Int): Int = { + var dp = new Array[Int](n + 1) + dp(0) = 1 + for (i <- 1 to n) { + for (j <- 1 to i) { + dp(i) += dp(j - 1) * dp(i - j) + } + } + dp(n) + } +} +``` + -----------------------
From cbaa9df25b535c7b1ff040bdc30631603e1cd3f7 Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Sat, 18 Jun 2022 16:58:58 -0500 Subject: [PATCH 079/136] =?UTF-8?q?=E4=BC=98=E5=8C=960925.=E9=95=BF?= =?UTF-8?q?=E6=8C=89=E9=94=AE=E5=85=A5python=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0925.长按键入.md | 38 +++++++++++++++----------------------- 1 file changed, 15 insertions(+), 23 deletions(-) diff --git a/problems/0925.长按键入.md b/problems/0925.长按键入.md index 0ef5a3d7..7aab71a2 100644 --- a/problems/0925.长按键入.md +++ b/problems/0925.长按键入.md @@ -129,29 +129,21 @@ class Solution { ``` ### Python ```python -class Solution: - def isLongPressedName(self, name: str, typed: str) -> bool: - i, j = 0, 0 - m, n = len(name) , len(typed) - while i< m and j < n: - if name[i] == typed[j]: # 相同时向后匹配 - i += 1 - j += 1 - else: # 不相同 - if j == 0: return False # 如果第一位不相同，直接返回false - # 判断边界为n-1,若为n会越界,例如name:"kikcxmvzi" typed:"kiikcxxmmvvzzz" - while j < n - 1 and typed[j] == typed[j-1]: j += 1 - if name[i] == typed[j]: - i += 1 - j += 1 - else: return False - # 说明name没有匹配完 - if i < m: return False - # 说明type没有匹配完 - while j < n: - if typed[j] == typed[j-1]: j += 1 - else: return False - return True + i = j = 0 + while(i Date: Sun, 19 Jun 2022 21:20:49 +0800 Subject: [PATCH 080/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E8=83=8C=E5=8C=85?= =?UTF-8?q?=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-1.md?= =?UTF-8?q?=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-1.md | 34 ++++++++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index a40a92ab..f9916667 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -498,7 +498,41 @@ const size = 4; console.log(testWeightBagProblem(weight, value, size)); ``` +### Scala +```scala +object Solution { + // 01背包 + def test_2_wei_bag_problem1(): Unit = { + var weight = Array[Int](1, 3, 4) + var value = Array[Int](15, 20, 30) + var baseweight = 4 + + // 二维数组 + var dp = Array.ofDim[Int](weight.length, baseweight + 1) + + // 初始化 + for (j <- weight(0) to baseweight) { + dp(0)(j) = value(0) + } + + // 遍历 + for (i <- 1 until weight.length; j <- 1 to baseweight) { + if (j - weight(i) >= 0) dp(i)(j) = dp(i - 1)(j - weight(i)) + value(i) + dp(i)(j) = math.max(dp(i)(j), dp(i - 1)(j)) + } + + // 打印数组 + dp.foreach(x => println("[" + x.mkString(",") + "]")) + + dp(weight.length - 1)(baseweight) // 最终返回 + } + + def main(args: Array[String]): Unit = { + test_2_wei_bag_problem1() + } +} +``` -----------------------
From a2d340b00279aac08df7cbba7a5733416dc28920 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sun, 19 Jun 2022 21:32:10 +0800 Subject: [PATCH 081/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E8=83=8C=E5=8C=85?= =?UTF-8?q?=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-2.md?= =?UTF-8?q?=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-2.md | 26 ++++++++++++++++++++++++++ 1 file changed, 26 insertions(+) diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index b66b74a6..81e61be4 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -375,7 +375,33 @@ console.log(testWeightBagProblem(weight, value, size)); ``` +### Scala +```scala +object Solution { + // 滚动数组 + def test_1_wei_bag_problem(): Unit = { + var weight = Array[Int](1, 3, 4) + var value = Array[Int](15, 20, 30) + var baseweight = 4 + + // dp数组 + var dp = new Array[Int](baseweight + 1) + + // 遍历 + for (i <- 0 until weight.length; j <- baseweight to weight(i) by -1) { + dp(j) = math.max(dp(j), dp(j - weight(i)) + value(i)) + } + + // 打印数组 + println("[" + dp.mkString(",") + "]") + } + + def main(args: Array[String]): Unit = { + test_1_wei_bag_problem() + } +} +``` -----------------------
From 1739379b3d31dfe41b519ee4f2eca3da03e53b79 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sun, 19 Jun 2022 21:57:51 +0800 Subject: [PATCH 082/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200416.=E5=88=86?= =?UTF-8?q?=E5=89=B2=E7=AD=89=E5=92=8C=E5=AD=90=E9=9B=86.md=20Scala?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0416.分割等和子集.md | 55 +++++++++++++++++++++++++++++++---- 1 file changed, 49 insertions(+), 6 deletions(-) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index eb6601e1..e14287e6 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -183,7 +183,7 @@ public: ## 其他语言版本 -Java： +### Java： ```Java class Solution { public boolean canPartition(int[] nums) { @@ -316,7 +316,7 @@ class Solution { } } ``` -Python： +### Python： ```python class Solution: def canPartition(self, nums: List[int]) -> bool: @@ -329,7 +329,7 @@ class Solution: dp[j] = max(dp[j], dp[j - nums[i]] + nums[i]) return target == dp[target] ``` -Go： +### Go： ```go // 分割等和子集 动态规划 // 时间复杂度O(n^2) 空间复杂度O(n) @@ -397,7 +397,7 @@ func canPartition(nums []int) bool { } ``` -javaScript: +### javaScript: ```js var canPartition = function(nums) { @@ -417,7 +417,7 @@ var canPartition = function(nums) { ``` -C: +### C: 二维dp： ```c /** @@ -518,7 +518,7 @@ bool canPartition(int* nums, int numsSize){ } ``` -TypeScript: +### TypeScript: > 一维数组，简洁 @@ -573,7 +573,50 @@ function canPartition(nums: number[]): boolean { }; ``` +### Scala +滚动数组: +```scala +object Solution { + def canPartition(nums: Array[Int]): Boolean = { + var sum = nums.sum + if (sum % 2 != 0) return false + var half = sum / 2 + var dp = new Array[Int](half + 1) + + // 遍历 + for (i <- 0 until nums.length; j <- half to nums(i) by -1) { + dp(j) = math.max(dp(j), dp(j - nums(i)) + nums(i)) + } + + if (dp(half) == half) true else false + } +} +``` + +二维数组: +```scala +object Solution { + def canPartition(nums: Array[Int]): Boolean = { + var sum = nums.sum + if (sum % 2 != 0) return false + var half = sum / 2 + var dp = Array.ofDim[Int](nums.length, half + 1) + + // 初始化 + for (j <- nums(0) to half) dp(0)(j) = nums(0) + + // 遍历 + for (i <- 1 until nums.length; j <- 1 to half) { + if (j - nums(i) >= 0) dp(i)(j) = nums(i) + dp(i - 1)(j - nums(i)) + dp(i)(j) = math.max(dp(i)(j), dp(i - 1)(j)) + } + + // 如果等于half就返回ture，否则返回false + if (dp(nums.length - 1)(half) == half) true else false + } +} +``` ----------------------- From 910dc88d54a098b374b7c137873e3711710d68f9 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Sun, 19 Jun 2022 22:50:50 +0800 Subject: [PATCH 083/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201049.=E6=9C=80?= =?UTF-8?q?=E5=90=8E=E4=B8=80=E5=9D=97=E7=9F=B3=E5=A4=B4=E7=9A=84=E9=87=8D?= =?UTF-8?q?=E9=87=8FII.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1049.最后一块石头的重量II.md | 50 ++++++++++++++++++++++++--- 1 file changed, 45 insertions(+), 5 deletions(-) diff --git a/problems/1049.最后一块石头的重量II.md b/problems/1049.最后一块石头的重量II.md index 3d256c3d..1e3b958f 100644 --- a/problems/1049.最后一块石头的重量II.md +++ b/problems/1049.最后一块石头的重量II.md @@ -152,7 +152,7 @@ public: ## 其他语言版本 -Java： +### Java： 一维数组版本 ```Java @@ -212,7 +212,7 @@ class Solution { ``` -Python： +### Python： ```python class Solution: def lastStoneWeightII(self, stones: List[int]) -> int: @@ -225,7 +225,7 @@ class Solution: return sumweight - 2 * dp[target] ``` -Go： +### Go： ```go func lastStoneWeightII(stones []int) int { // 15001 = 30 * 1000 /2 +1 @@ -254,7 +254,7 @@ func max(a, b int) int { } ``` -JavaScript版本 +### JavaScript ```javascript /** @@ -277,7 +277,7 @@ var lastStoneWeightII = function (stones) { }; ``` -TypeScript： +### TypeScript： ```typescript function lastStoneWeightII(stones: number[]): number { @@ -296,7 +296,47 @@ function lastStoneWeightII(stones: number[]): number { }; ``` +### Scala +滚动数组: +```scala +object Solution { + def lastStoneWeightII(stones: Array[Int]): Int = { + var sum = stones.sum + var half = sum / 2 + var dp = new Array[Int](half + 1) + + // 遍历 + for (i <- 0 until stones.length; j <- half to stones(i) by -1) { + dp(j) = math.max(dp(j), dp(j - stones(i)) + stones(i)) + } + + sum - 2 * dp(half) + } +} +``` + +二维数组: +```scala +object Solution { + def lastStoneWeightII(stones: Array[Int]): Int = { + var sum = stones.sum + var half = sum / 2 + var dp = Array.ofDim[Int](stones.length, half + 1) + + // 初始化 + for (j <- stones(0) to half) dp(0)(j) = stones(0) + + // 遍历 + for (i <- 1 until stones.length; j <- 1 to half) { + if (j - stones(i) >= 0) dp(i)(j) = stones(i) + dp(i - 1)(j - stones(i)) + dp(i)(j) = math.max(dp(i)(j), dp(i - 1)(j)) + } + + sum - 2 * dp(stones.length - 1)(half) + } +} +``` -----------------------
From 4cb3897549197115721472552bbfb46e3848d682 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Mon, 20 Jun 2022 21:19:31 +0800 Subject: [PATCH 084/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200494.=E7=9B=AE?= =?UTF-8?q?=E6=A0=87=E5=92=8C.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0494.目标和.md | 28 +++++++++++++++++++++++----- 1 file changed, 23 insertions(+), 5 deletions(-) diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 8ce1f6f1..639f8ee8 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -250,7 +250,7 @@ dp[j] += dp[j - nums[i]]; ## 其他语言版本 -Java： +### Java ```java class Solution { public int findTargetSumWays(int[] nums, int target) { @@ -271,7 +271,7 @@ class Solution { } ``` -Python： +### Python ```python class Solution: def findTargetSumWays(self, nums: List[int], target: int) -> int: @@ -287,7 +287,7 @@ class Solution: return dp[bagSize] ``` -Go： +### Go ```go func findTargetSumWays(nums []int, target int) int { sum := 0 @@ -322,7 +322,7 @@ func abs(x int) int { } ``` -Javascript： +### Javascript ```javascript const findTargetSumWays = (nums, target) => { @@ -351,7 +351,7 @@ const findTargetSumWays = (nums, target) => { }; ``` -TypeScript： +### TypeScript ```typescript function findTargetSumWays(nums: number[], target: number): number { @@ -370,7 +370,25 @@ function findTargetSumWays(nums: number[], target: number): number { }; ``` +### Scala +```scala +object Solution { + def findTargetSumWays(nums: Array[Int], target: Int): Int = { + var sum = nums.sum + if (math.abs(target) > sum) return 0 // 此时没有方案 + if ((sum + target) % 2 == 1) return 0 // 此时没有方案 + var bagSize = (sum + target) / 2 + var dp = new Array[Int](bagSize + 1) + dp(0) = 1 + for (i <- 0 until nums.length; j <- bagSize to nums(i) by -1) { + dp(j) += dp(j - nums(i)) + } + + dp(bagSize) + } +} +``` -----------------------
From 2d9727d6202f802674c3cb399f837b66c05b6350 Mon Sep 17 00:00:00 2001 From: Parker999 Date: Mon, 20 Jun 2022 18:35:43 -0700 Subject: [PATCH 085/136] update the dp494 --- problems/0494.目标和.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 99b76834..60f721c2 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -213,6 +213,7 @@ public: if (abs(S) > sum) return 0; // 此时没有方案 if ((S + sum) % 2 == 1) return 0; // 此时没有方案 int bagSize = (S + sum) / 2; + if(bagsize<0) return 0; vector dp(bagSize + 1, 0); dp[0] = 1; for (int i = 0; i < nums.size(); i++) { From 89c9044bf984ad638d07932f3b9ebe91e2e7589d Mon Sep 17 00:00:00 2001 From: xiaojun <13589818805@163.com> Date: Thu, 23 Jun 2022 15:36:55 +0800 Subject: [PATCH 086/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=881382.=20?= =?UTF-8?q?=E5=B0=86=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E5=8F=98?= =?UTF-8?q?=E5=B9=B3=E8=A1=A1=EF=BC=89=E7=9A=84go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1382.将二叉搜索树变平衡.md | 40 +++++++++++++++++++++++++++++ 1 file changed, 40 insertions(+) diff --git a/problems/1382.将二叉搜索树变平衡.md b/problems/1382.将二叉搜索树变平衡.md index 57231ec4..7c7f8484 100644 --- a/problems/1382.将二叉搜索树变平衡.md +++ b/problems/1382.将二叉搜索树变平衡.md @@ -123,6 +123,46 @@ class Solution: ``` Go： +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func balanceBST(root *TreeNode) *TreeNode { + // 二叉搜索树中序遍历得到有序数组 + nums := []int{} + // 中序递归遍历二叉树 + var travel func(node *TreeNode) + travel = func(node *TreeNode) { + if node == nil { + return + } + travel(node.Left) + nums = append(nums, node.Val) + travel(node.Right) + } + // 二分法保证左右子树高度差不超过一（题目要求返回的仍是二叉搜索树） + var buildTree func(nums []int, left, right int) *TreeNode + buildTree = func(nums []int, left, right int) *TreeNode { + if left > right { + return nil + } + mid := left + (right-left) >> 1 + root := &TreeNode{Val: nums[mid]} + root.Left = buildTree(nums, left, mid-1) + root.Right = buildTree(nums, mid+1, right) + return root + } + travel(root) + return buildTree(nums, 0, len(nums)-1) +} + +``` + JavaScript： ```javascript var balanceBST = function(root) { From beb6805c1d7fd77b66ba0870870fd12228d5a4cf Mon Sep 17 00:00:00 2001 From: unknown Date: Thu, 23 Jun 2022 10:29:24 +0100 Subject: [PATCH 087/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201049.=E6=9C=80?= =?UTF-8?q?=E5=90=8E=E4=B8=80=E5=9D=97=E7=9F=B3=E5=A4=B4=E7=9A=84=E9=87=8D?= =?UTF-8?q?=E9=87=8F.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1049.最后一块石头的重量II.md | 31 +++++++++++++++++++++++++++ 1 file changed, 31 insertions(+) diff --git a/problems/1049.最后一块石头的重量II.md b/problems/1049.最后一块石头的重量II.md index ee0ddef2..c49f0a18 100644 --- a/problems/1049.最后一块石头的重量II.md +++ b/problems/1049.最后一块石头的重量II.md @@ -277,5 +277,36 @@ var lastStoneWeightII = function (stones) { }; ``` +C版本 +```c +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) + +int getSum(int *stones, int stoneSize) { + int sum = 0, i; + for (i = 0; i < stoneSize; ++i) + sum += stones[i]; + return sum; +} + +int lastStoneWeightII(int* stones, int stonesSize){ + int sum = getSum(stones, stonesSize); + int target = sum / 2; + int i, j; + + // 初始化dp数组 + int *dp = (int*)malloc(sizeof(int) * (target + 1)); + memset(dp, 0, sizeof(int) * (target + 1)); + for (j = stones[0]; j <= target; ++j) + dp[j] = stones[0]; + + // 递推公式：dp[j] = max(dp[j], dp[j - stones[i]] + stones[i]) + for (i = 1; i < stonesSize; ++i) { + for (j = target; j >= stones[i]; --j) + dp[j] = MAX(dp[j], dp[j - stones[i]] + stones[i]); + } + return sum - dp[target] - dp[target]; +} +``` + -----------------------
From c99bf39601a06278f9cca2e5f6d61638e23f3ced Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Thu, 23 Jun 2022 21:34:24 +0800 Subject: [PATCH 088/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200474.=E4=B8=80?= =?UTF-8?q?=E5=92=8C=E9=9B=B6.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0474.一和零.md | 83 ++++++++++++++++++++++++++++++++++++++--- 1 file changed, 78 insertions(+), 5 deletions(-) diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md index d38ce03f..d6c598aa 100644 --- a/problems/0474.一和零.md +++ b/problems/0474.一和零.md @@ -163,7 +163,7 @@ public: ## 其他语言版本 -Java： +### Java ```Java class Solution { public int findMaxForm(String[] strs, int m, int n) { @@ -192,7 +192,7 @@ class Solution { } ``` -Python： +### Python ```python class Solution: def findMaxForm(self, strs: List[str], m: int, n: int) -> int: @@ -208,7 +208,7 @@ class Solution: return dp[m][n] ``` -Go： +### Go ```go func findMaxForm(strs []string, m int, n int) int { // 定义数组 @@ -294,7 +294,7 @@ func getMax(a,b int)int{ } ``` -Javascript： +### Javascript ```javascript const findMaxForm = (strs, m, n) => { const dp = Array.from(Array(m+1), () => Array(n+1).fill(0)); @@ -323,7 +323,7 @@ const findMaxForm = (strs, m, n) => { }; ``` -TypeScript： +### TypeScript > 滚动数组，二维数组法 @@ -446,7 +446,80 @@ function isValidSubSet(strs: string[], m: number, n: number): boolean { } ``` +### Scala +背包： +```scala +object Solution { + def findMaxForm(strs: Array[String], m: Int, n: Int): Int = { + var dp = Array.ofDim[Int](m + 1, n + 1) + + var (oneNum, zeroNum) = (0, 0) + + for (str <- strs) { + oneNum = 0 + zeroNum = 0 + for (i <- str.indices) { + if (str(i) == '0') zeroNum += 1 + else oneNum += 1 + } + + for (i <- m to zeroNum by -1) { + for (j <- n to oneNum by -1) { + dp(i)(j) = math.max(dp(i)(j), dp(i - zeroNum)(j - oneNum) + 1) + } + } + } + + dp(m)(n) + } +} +``` + +回溯法(超时)： +```scala +object Solution { + import scala.collection.mutable + + var res = Int.MinValue + + def test(str: String): (Int, Int) = { + var (zero, one) = (0, 0) + for (i <- str.indices) { + if (str(i) == '1') one += 1 + else zero += 1 + } + (zero, one) + } + + def travsel(strs: Array[String], path: mutable.ArrayBuffer[String], m: Int, n: Int, startIndex: Int): Unit = { + if (startIndex > strs.length) { + return + } + + res = math.max(res, path.length) + + for (i <- startIndex until strs.length) { + + var (zero, one) = test(strs(i)) + + // 如果0的个数小于m，1的个数小于n，则可以回溯 + if (zero <= m && one <= n) { + path.append(strs(i)) + travsel(strs, path, m - zero, n - one, i + 1) + path.remove(path.length - 1) + } + } + } + + def findMaxForm(strs: Array[String], m: Int, n: Int): Int = { + res = Int.MinValue + var path = mutable.ArrayBuffer[String]() + travsel(strs, path, m, n, 0) + res + } +} +``` -----------------------
From e9edda44e0cf080701614a1df8ab680a264c4e40 Mon Sep 17 00:00:00 2001 From: shutengfei Date: Sat, 25 Jun 2022 15:56:26 +0800 Subject: [PATCH 089/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880649.Dota2?= =?UTF-8?q?=E5=8F=82=E8=AE=AE=E9=99=A2.md=EF=BC=89=EF=BC=9A=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0649.Dota2参议院.md | 38 ++++++++++++++++++++++++++++++++++++ 1 file changed, 38 insertions(+) diff --git a/problems/0649.Dota2参议院.md b/problems/0649.Dota2参议院.md index 6e84c9fd..264a003a 100644 --- a/problems/0649.Dota2参议院.md +++ b/problems/0649.Dota2参议院.md @@ -244,6 +244,44 @@ var predictPartyVictory = function(senateStr) { }; ``` +## TypeScript + +```typescript +function predictPartyVictory(senate: string): string { + // 数量差：Count(Radiant) - Count(Dire) + let deltaRDCnt: number = 0; + let hasR: boolean = true, + hasD: boolean = true; + const senateArr: string[] = senate.split(''); + while (hasR && hasD) { + hasR = false; + hasD = false; + for (let i = 0, length = senateArr.length; i < length; i++) { + if (senateArr[i] === 'R') { + if (deltaRDCnt < 0) { + senateArr[i] = ''; + } else { + hasR = true; + } + deltaRDCnt++; + } else if (senateArr[i] === 'D') { + if (deltaRDCnt > 0) { + senateArr[i] = ''; + } else { + hasD = true; + } + deltaRDCnt--; + } + } + } + return hasR ? 'Radiant' : 'Dire'; +}; +``` + + + + + -----------------------
From dd31d67778cea30688bd60e4e0b23631758538e8 Mon Sep 17 00:00:00 2001 From: lesenelir Date: Sun, 26 Jun 2022 11:24:35 +0800 Subject: [PATCH 090/136] =?UTF-8?q?=E4=BF=AE=E6=94=B9=EF=BC=9A(1049.?= =?UTF-8?q?=E6=9C=80=E5=90=8E=E4=B8=80=E5=9D=97=E7=9F=B3=E5=A4=B4=E7=9A=84?= =?UTF-8?q?=E9=87=8D=E9=87=8FII.md):=20=E4=BF=AE=E6=94=B9markdown=E6=96=87?= =?UTF-8?q?=E6=A1=A3?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1049.最后一块石头的重量II.md | 1 - 1 file changed, 1 deletion(-) diff --git a/problems/1049.最后一块石头的重量II.md b/problems/1049.最后一块石头的重量II.md index f3e7909c..1e87848e 100644 --- a/problems/1049.最后一块石头的重量II.md +++ b/problems/1049.最后一块石头的重量II.md @@ -3,7 +3,6 @@

参与本项目，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！

-# 动态规划：最后一块石头的重量 II ## 1049. 最后一块石头的重量 II From 88ca27562e748e004ac75dfd9ad89c2fc8fc7f07 Mon Sep 17 00:00:00 2001 From: Yang Date: Sun, 26 Jun 2022 10:08:22 +0200 Subject: [PATCH 091/136] Remove redundant `break` in the for loop MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Remove redundant `break` in the for loop of python script. When the `return` is triggered, the for loop will break automatically. PS: really like your work! Thanks a lot 😄 . --- problems/0242.有效的字母异位词.md | 2 -- 1 file changed, 2 deletions(-) diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index 8fd9c604..f1f7e6cf 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -125,8 +125,6 @@ class Solution: if record[i] != 0: #record数组如果有的元素不为零0，说明字符串s和t 一定是谁多了字符或者谁少了字符。 return False - #如果有一个元素不为零，则可以判断字符串s和t不是字母异位词 - break return True ``` From ffe981fb6c5af5c7400f3b82c455b80ac8333fc0 Mon Sep 17 00:00:00 2001 From: AronJudge <2286381138@qq.com> Date: Sun, 26 Jun 2022 23:50:32 +0800 Subject: [PATCH 092/136] =?UTF-8?q?0977.=20=E6=9C=89=E5=BA=8F=E6=95=B0?= =?UTF-8?q?=E7=BB=84=E7=9A=84=E5=B9=B3=E6=96=B9=20=E6=B7=BB=E5=8A=A0Java?= =?UTF-8?q?=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0977.有序数组的平方.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index 20bdf7b0..d274d778 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -106,6 +106,7 @@ class Solution { int index = result.length - 1; while (left <= right) { if (nums[left] * nums[left] > nums[right] * nums[right]) { + // 正数的相对位置是不变的， 需要调整的是负数平方后的相对位置 result[index--] = nums[left] * nums[left]; ++left; } else { From 8bc026fed035ee9c0f2f8b0452f7e0a5f5531281 Mon Sep 17 00:00:00 2001 From: UndeadSheep Date: Mon, 27 Jun 2022 15:28:47 +0800 Subject: [PATCH 093/136] =?UTF-8?q?=E6=96=B0=E5=A2=9E=20=E6=A0=88=E4=B8=8E?= =?UTF-8?q?=E9=98=9F=E5=88=97=E9=83=A8=E5=88=86=E7=9A=84=20C#=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0020.有效的括号.md | 31 ++++++++++++++ problems/0150.逆波兰表达式求值.md | 34 +++++++++++++++ problems/0225.用队列实现栈.md | 35 ++++++++++++++++ problems/0232.用栈实现队列.md | 41 +++++++++++++++++++ problems/0239.滑动窗口最大值.md | 41 +++++++++++++++++++ problems/0347.前K个高频元素.md | 35 ++++++++++++++++ problems/1047.删除字符串中的所有相邻重复项.md | 17 ++++++++ 7 files changed, 234 insertions(+) diff --git a/problems/0020.有效的括号.md b/problems/0020.有效的括号.md index 7bb7f746..cdca70f9 100644 --- a/problems/0020.有效的括号.md +++ b/problems/0020.有效的括号.md @@ -401,5 +401,36 @@ bool isValid(char * s){ } ``` +C#: +```csharp +public class Solution { + public bool IsValid(string s) { + var len = s.Length; + if(len % 2 == 1) return false; // 字符串长度为单数，直接返回 false + // 初始化栈 + var stack = new Stack(); + // 遍历字符串 + for(int i = 0; i < len; i++){ + // 当字符串为左括号时，进栈对应的右括号 + if(s[i] == '('){ + stack.Push(')'); + }else if(s[i] == '['){ + stack.Push(']'); + }else if(s[i] == '{'){ + stack.Push('}'); + } + // 当字符串为右括号时，当栈为空（无左括号） 或者 出栈字符不是当前的字符 + else if(stack.Count == 0 || stack.Pop() != s[i]) + return false; + } + // 如果栈不为空，例如“((()”，右括号少于左括号，返回false + if (stack.Count > 0) + return false; + // 上面的校验都满足，则返回true + else + return true; + } +} +``` -----------------------
diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md index f4dad823..8dafa460 100644 --- a/problems/0150.逆波兰表达式求值.md +++ b/problems/0150.逆波兰表达式求值.md @@ -326,5 +326,39 @@ func evalRPN(_ tokens: [String]) -> Int { } ``` +C#: +```csharp +public int EvalRPN(string[] tokens) { + int num; + Stack stack = new Stack(); + foreach(string s in tokens){ + if(int.TryParse(s, out num)){ + stack.Push(num); + }else{ + int num1 = stack.Pop(); + int num2 = stack.Pop(); + switch (s) + { + case "+": + stack.Push(num1 + num2); + break; + case "-": + stack.Push(num2 - num1); + break; + case "*": + stack.Push(num1 * num2); + break; + case "/": + stack.Push(num2 / num1); + break; + default: + break; + } + } + } + return stack.Pop(); + } +``` + -----------------------
diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index 3457c4b3..5d902fb1 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -816,5 +816,40 @@ class MyStack { } ``` +C#: +```csharp +public class MyStack { + Queue queue1; + Queue queue2; + public MyStack() { + queue1 = new Queue(); + queue2 = new Queue(); + } + + public void Push(int x) { + queue2.Enqueue(x); + while(queue1.Count != 0){ + queue2.Enqueue(queue1.Dequeue()); + } + Queue queueTemp; + queueTemp = queue1; + queue1 = queue2; + queue2 = queueTemp; + } + + public int Pop() { + return queue1.Count > 0 ? queue1.Dequeue() : -1; + } + + public int Top() { + return queue1.Count > 0 ? queue1.Peek() : -1; + } + + public bool Empty() { + return queue1.Count == 0; + } +} + +``` -----------------------
diff --git a/problems/0232.用栈实现队列.md b/problems/0232.用栈实现队列.md index 1a56d9f3..d58dc55f 100644 --- a/problems/0232.用栈实现队列.md +++ b/problems/0232.用栈实现队列.md @@ -496,5 +496,46 @@ void myQueueFree(MyQueue* obj) { } ``` +C#: +```csharp +public class MyQueue { + Stack inStack; + Stack outStack; + + public MyQueue() { + inStack = new Stack();// 负责进栈 + outStack = new Stack();// 负责出栈 + } + + public void Push(int x) { + inStack.Push(x); + } + + public int Pop() { + dumpstackIn(); + return outStack.Pop(); + } + + public int Peek() { + dumpstackIn(); + return outStack.Peek(); + } + + public bool Empty() { + return inStack.Count == 0 && outStack.Count == 0; + } + + // 处理方法： + // 如果outStack为空，那么将inStack中的元素全部放到outStack中 + private void dumpstackIn(){ + if (outStack.Count != 0) return; + while(inStack.Count != 0){ + outStack.Push(inStack.Pop()); + } + } +} + +``` + -----------------------
diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md index f269450f..35ca1eed 100644 --- a/problems/0239.滑动窗口最大值.md +++ b/problems/0239.滑动窗口最大值.md @@ -631,5 +631,46 @@ func maxSlidingWindow(_ nums: [Int], _ k: Int) -> [Int] { } ``` +C#: +```csharp +class myDequeue{ + private LinkedList linkedList = new LinkedList(); + + public void Enqueue(int n){ + while(linkedList.Count > 0 && linkedList.Last.Value < n){ + linkedList.RemoveLast(); + } + linkedList.AddLast(n); + } + + public int Max(){ + return linkedList.First.Value; + } + + public void Dequeue(int n){ + if(linkedList.First.Value == n){ + linkedList.RemoveFirst(); + } + } + } + + myDequeue window = new myDequeue(); + List res = new List(); + + public int[] MaxSlidingWindow(int[] nums, int k) { + for(int i = 0; i < k; i++){ + window.Enqueue(nums[i]); + } + res.Add(window.Max()); + for(int i = k; i < nums.Length; i++){ + window.Dequeue(nums[i-k]); + window.Enqueue(nums[i]); + res.Add(window.Max()); + } + + return res.ToArray(); + } +``` + -----------------------
diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md index 1d6a358b..c570672f 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -374,7 +374,42 @@ function topKFrequent(nums: number[], k: number): number[] { }; ``` +C#: +```csharp + public int[] TopKFrequent(int[] nums, int k) { + //哈希表-标权重 + Dictionary dic = new(); + for(int i = 0; i < nums.Length; i++){ + if(dic.ContainsKey(nums[i])){ + dic[nums[i]]++; + }else{ + dic.Add(nums[i], 1); + } + } + //优先队列-从小到大排列 + PriorityQueue pq = new(); + foreach(var num in dic){ + pq.Enqueue(num.Key, num.Value); + if(pq.Count > k){ + pq.Dequeue(); + } + } + + // //Stack-倒置 + // Stack res = new(); + // while(pq.Count > 0){ + // res.Push(pq.Dequeue()); + // } + // return res.ToArray(); + //数组倒装 + int[] res = new int[k]; + for(int i = k - 1; i >= 0; i--){ + res[i] = pq.Dequeue(); + } + return res; + } +``` -----------------------
diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md index 638c8f4e..9237acdb 100644 --- a/problems/1047.删除字符串中的所有相邻重复项.md +++ b/problems/1047.删除字符串中的所有相邻重复项.md @@ -375,5 +375,22 @@ func removeDuplicates(_ s: String) -> String { } ``` +C#: +```csharp +public string RemoveDuplicates(string s) { + //拿字符串直接作为栈，省去了栈还要转为字符串的操作 + StringBuilder res = new StringBuilder(); + + foreach(char c in s){ + if(res.Length > 0 && res[res.Length-1] == c){ + res.Remove(res.Length-1, 1); + }else{ + res.Append(c); + } + } + + return res.ToString(); + } +``` -----------------------
From 16af63e22ad0d3522c19ccf5eb3dd4a0b4294a9e Mon Sep 17 00:00:00 2001 From: Shuo Zhang Date: Mon, 27 Jun 2022 22:54:52 -0400 Subject: [PATCH 094/136] =?UTF-8?q?=E4=BF=AE=E6=94=B90349=20-=20=E7=94=A8?= =?UTF-8?q?=20Java=20stream=20=E4=BB=A3=E6=9B=BF=20for=20loop=20(=E8=BF=99?= =?UTF-8?q?=E6=A0=B7=E4=B8=BB=E9=A2=98=E6=9B=B4=E6=98=8E=E6=98=BE=EF=BC=89?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0349.两个数组的交集.md | 7 +------ 1 file changed, 1 insertion(+), 6 deletions(-) diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 98518647..2a8b2dae 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -104,13 +104,8 @@ class Solution { resSet.add(i); } } - int[] resArr = new int[resSet.size()]; - int index = 0; //将结果几何转为数组 - for (int i : resSet) { - resArr[index++] = i; - } - return resArr; + return resSet.stream().mapToInt(x -> x).toArray(); } } ``` From c6db9037901601bd84ed907455cbe2adf9f2e9b9 Mon Sep 17 00:00:00 2001 From: White Dove <43168562+zhouweiwei18@users.noreply.github.com> Date: Tue, 28 Jun 2022 16:24:27 +0800 Subject: [PATCH 095/136] =?UTF-8?q?Update=200494.=E7=9B=AE=E6=A0=87?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 细节 --- problems/0494.目标和.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 8f116fae..00e4bdfa 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -156,7 +156,7 @@ dp[j] 表示：填满j（包括j）这么大容积的包，有dp[j]种方法 有哪些来源可以推出dp[j]呢？ -不考虑nums[i]的情况下，填满容量为j - nums[i]的背包，有dp[j - nums[i]]种方法。 +不考虑nums[i]的情况下，填满容量为j的背包，有dp[j]种方法。 那么只要搞到nums[i]的话，凑成dp[j]就有dp[j - nums[i]] 种方法。 From c70316d7aa59caffe1a0cec9b1c818fcc867a411 Mon Sep 17 00:00:00 2001 From: wzqwtt <1722249371@qq.com> Date: Tue, 28 Jun 2022 21:30:49 +0800 Subject: [PATCH 096/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E8=83=8C=E5=8C=85?= =?UTF-8?q?=E9=97=AE=E9=A2=98=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80=E5=AE=8C?= =?UTF-8?q?=E5=85=A8=E8=83=8C=E5=8C=85.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包问题理论基础完全背包.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/problems/背包问题理论基础完全背包.md b/problems/背包问题理论基础完全背包.md index 54e772e0..fc4609a6 100644 --- a/problems/背包问题理论基础完全背包.md +++ b/problems/背包问题理论基础完全背包.md @@ -359,7 +359,27 @@ function test_CompletePack(): void { test_CompletePack(); ``` +Scala: +```scala +// 先遍历物品，再遍历背包容量 +object Solution { + def test_CompletePack() { + var weight = Array[Int](1, 3, 4) + var value = Array[Int](15, 20, 30) + var baseweight = 4 + + var dp = new Array[Int](baseweight + 1) + + for (i <- 0 until weight.length) { + for (j <- weight(i) to baseweight) { + dp(j) = math.max(dp(j), dp(j - weight(i)) + value(i)) + } + } + dp(baseweight) + } +} +``` ----------------------- From bdd4c83b642dfa5b2c7c5a6b6fd214aa8e0488bb Mon Sep 17 00:00:00 2001 From: wzqwtt <1722249371@qq.com> Date: Tue, 28 Jun 2022 21:56:42 +0800 Subject: [PATCH 097/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200518.=E9=9B=B6?= =?UTF-8?q?=E9=92=B1=E5=85=91=E6=8D=A2II.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0518.零钱兑换II.md | 15 +++++++++++++++ 1 file changed, 15 insertions(+) diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md index 222a10d7..3abb9601 100644 --- a/problems/0518.零钱兑换II.md +++ b/problems/0518.零钱兑换II.md @@ -289,7 +289,22 @@ function change(amount: number, coins: number[]): number { }; ``` +Scala: +```scala +object Solution { + def change(amount: Int, coins: Array[Int]): Int = { + var dp = new Array[Int](amount + 1) + dp(0) = 1 + for (i <- 0 until coins.length) { + for (j <- coins(i) to amount) { + dp(j) += dp(j - coins(i)) + } + } + dp(amount) + } +} +``` -----------------------
From 0cd4ffd9a470038f1b77207af082a47c00e8d430 Mon Sep 17 00:00:00 2001 From: azou Date: Wed, 29 Jun 2022 23:32:29 +0800 Subject: [PATCH 098/136] =?UTF-8?q?=E4=BF=AE=E6=AD=A3=EF=BC=9A=E7=A7=BB?= =?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0TS=E7=89=88?= =?UTF-8?q?=E6=9C=AC=E4=BB=A3=E7=A0=81=E9=94=99=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0203.移除链表元素.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index b79d29a5..0e461ce8 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -397,18 +397,18 @@ function removeElements(head: ListNode | null, val: number): ListNode | null { ```typescript function removeElements(head: ListNode | null, val: number): ListNode | null { - let dummyHead = new ListNode(0, head); - let pre: ListNode = dummyHead, cur: ListNode | null = dummyHead.next; - // 删除非头部节点 + // 添加虚拟节点 + const data = new ListNode(0, head); + let pre = data, cur = data.next; while (cur) { if (cur.val === val) { - pre.next = cur.next; + pre.next = cur.next } else { pre = cur; } cur = cur.next; } - return head.next; + return data.next; }; ``` From 55086c231acbb8794299d2f143444173de2b85e9 Mon Sep 17 00:00:00 2001 From: w2xi <43wangxi@gmail.com> Date: Sat, 2 Jul 2022 11:14:57 +0800 Subject: [PATCH 099/136] =?UTF-8?q?Update=200226.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md=20JavaScript=E9=80=92=E5=BD=92?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0226.翻转二叉树.md | 25 +++++++------------------ 1 file changed, 7 insertions(+), 18 deletions(-) diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md index 8e35fc9d..83d20df8 100644 --- a/problems/0226.翻转二叉树.md +++ b/problems/0226.翻转二叉树.md @@ -470,25 +470,14 @@ func invertTree(root *TreeNode) *TreeNode { 使用递归版本的前序遍历 ```javascript var invertTree = function(root) { - //1. 首先使用递归版本的前序遍历实现二叉树翻转 - //交换节点函数 - const inverNode=function(left,right){ - let temp=left; - left=right; - right=temp; - //需要重新给root赋值一下 - root.left=left; - root.right=right; + // 终止条件 + if (!root) { + return null; } - //确定递归函数的参数和返回值inverTree=function(root) - //确定终止条件 - if(root===null){ - return root; - } - //确定节点处理逻辑 交换 - inverNode(root.left,root.right); - invertTree(root.left); - invertTree(root.right); + // 交换左右节点 + const rightNode = root.right; + root.right = invertTree(root.left); + root.left = invertTree(rightNode); return root; }; ``` From 889256a3781ebc8891f8a37778334f402acf362e Mon Sep 17 00:00:00 2001 From: chenwingsing <742474834@qq.com> Date: Sat, 2 Jul 2022 21:58:17 +0800 Subject: [PATCH 100/136] =?UTF-8?q?=E6=9B=B4=E6=96=B00018JAVA=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 用例有一个是[1000000000,1000000000,1000000000,1000000000] -294967296 如果用int会越界，所以修改为long --- problems/0018.四数之和.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 2146a114..99613070 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -153,7 +153,7 @@ class Solution { int left = j + 1; int right = nums.length - 1; while (right > left) { - int sum = nums[i] + nums[j] + nums[left] + nums[right]; + long sum = (long) nums[i] + nums[j] + nums[left] + nums[right]; if (sum > target) { right--; } else if (sum < target) { From 1965ff69f70373fd48f31c590e77152444ceefe3 Mon Sep 17 00:00:00 2001 From: chenwingsing <742474834@qq.com> Date: Sun, 3 Jul 2022 09:07:51 +0800 Subject: [PATCH 101/136] =?UTF-8?q?=E6=9B=B4=E6=96=B0[0015.=E4=B8=89?= =?UTF-8?q?=E6=95=B0=E4=B9=8B=E5=92=8C]=E8=A7=A3=E9=A2=98=E6=8F=8F?= =?UTF-8?q?=E8=BF=B0=E9=83=A8=E5=88=86?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加逗号让解题更加清晰，避免歧义，我一开始就看成a = num[i] *b了 --- problems/0015.三数之和.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md index 4f1d711a..8bc8dd91 100644 --- a/problems/0015.三数之和.md +++ b/problems/0015.三数之和.md @@ -95,7 +95,7 @@ public: 拿这个nums数组来举例，首先将数组排序，然后有一层for循环，i从下标0的地方开始，同时定一个下标left 定义在i+1的位置上，定义下标right 在数组结尾的位置上。 -依然还是在数组中找到 abc 使得a + b +c =0，我们这里相当于 a = nums[i] b = nums[left] c = nums[right]。 +依然还是在数组中找到 abc 使得a + b +c =0，我们这里相当于 a = nums[i]，b = nums[left]，c = nums[right]。 接下来如何移动left 和right呢， 如果nums[i] + nums[left] + nums[right] > 0 就说明 此时三数之和大了，因为数组是排序后了，所以right下标就应该向左移动，这样才能让三数之和小一些。 From 8c54205566578bf331587de936e85b5bc4454e5a Mon Sep 17 00:00:00 2001 From: chenwingsing <742474834@qq.com> Date: Sun, 3 Jul 2022 11:53:38 +0800 Subject: [PATCH 102/136] =?UTF-8?q?Update=200151.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改错误字 --- problems/0151.翻转字符串里的单词.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index 38372f91..23f966e8 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -79,7 +79,7 @@ void removeExtraSpaces(string& s) { 逻辑很简单，从前向后遍历，遇到空格了就erase。 -如果不仔细琢磨一下erase的时间复杂读，还以为以上的代码是O(n)的时间复杂度呢。 +如果不仔细琢磨一下erase的时间复杂度，还以为以上的代码是O(n)的时间复杂度呢。 想一下真正的时间复杂度是多少，一个erase本来就是O(n)的操作，erase实现原理题目：[数组：就移除个元素很难么？](https://programmercarl.com/0027.移除元素.html)，最优的算法来移除元素也要O(n)。 From 11b701fdb6ba689bdc9f07362e497a4d2ab5eefd Mon Sep 17 00:00:00 2001 From: AronJudge <2286381138@qq.com> Date: Mon, 4 Jul 2022 00:58:56 +0800 Subject: [PATCH 103/136] =?UTF-8?q?0018=20Java=E4=BB=A3=E7=A0=81=E9=83=A8?= =?UTF-8?q?=E5=88=86=E5=A2=9E=E5=8A=A0=E5=89=AA=E6=9E=9D=E6=93=8D=E4=BD=9C?= =?UTF-8?q?,=E4=B8=8D=E7=84=B6Leetcode=E4=B8=8D=E8=83=BD=E9=80=9A=E8=BF=87?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0018.四数之和.md | 5 +++++ 1 file changed, 5 insertions(+) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 6cbd40c2..d0b7fc68 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -140,6 +140,11 @@ class Solution { for (int i = 0; i < nums.length; i++) { + // nums[i] > target 直接返回, 剪枝操作 + if (nums[i] > 0 && nums[i] > target) { + return result; + } + if (i > 0 && nums[i - 1] == nums[i]) { continue; } From ca00ec680557689d6b302fe27b3e66e479a0d421 Mon Sep 17 00:00:00 2001 From: xiaojun <13589818805@163.com> Date: Mon, 4 Jul 2022 15:43:35 +0800 Subject: [PATCH 104/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0(0031.=E4=B8=8B?= =?UTF-8?q?=E4=B8=80=E4=B8=AA=E6=8E=92=E5=88=97.md)go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0031.下一个排列.md | 24 ++++++++++++++++++++++-- 1 file changed, 22 insertions(+), 2 deletions(-) diff --git a/problems/0031.下一个排列.md b/problems/0031.下一个排列.md index bce8adef..1a3641b0 100644 --- a/problems/0031.下一个排列.md +++ b/problems/0031.下一个排列.md @@ -171,13 +171,13 @@ class Solution(object): i = n-2 while i >= 0 and nums[i] >= nums[i+1]: i -= 1 - + if i > -1: // i==-1,不存在下一个更大的排列 j = n-1 while j >= 0 and nums[j] <= nums[i]: j -= 1 nums[i], nums[j] = nums[j], nums[i] - + start, end = i+1, n-1 while start < end: nums[start], nums[end] = nums[end], nums[start] @@ -190,6 +190,26 @@ class Solution(object): ## Go ```go +//卡尔的解法 +func nextPermutation(nums []int) { + for i:=len(nums)-1;i>=0;i--{ + for j:=len(nums)-1;j>i;j--{ + if nums[j]>nums[i]{ + //交换 + nums[j],nums[i]=nums[i],nums[j] + reverse(nums,0+i+1,len(nums)-1) + return + } + } + } + reverse(nums,0,len(nums)-1) +} +//对目标切片指定区间的反转方法 +func reverse(a []int,begin,end int){ + for i,j:=begin,end;i Date: Wed, 6 Jul 2022 22:01:14 +0800 Subject: [PATCH 105/136] =?UTF-8?q?Update=200450.=E5=88=A0=E9=99=A4?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84?= =?UTF-8?q?=E8=8A=82=E7=82=B9.md=20JavaScript=E9=80=92=E5=BD=92=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0450.删除二叉搜索树中的节点.md | 57 +++++++++++++++---------- 1 file changed, 34 insertions(+), 23 deletions(-) diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md index 3fa2a1c5..3bc8369c 100644 --- a/problems/0450.删除二叉搜索树中的节点.md +++ b/problems/0450.删除二叉搜索树中的节点.md @@ -456,31 +456,42 @@ func deleteNode(root *TreeNode, key int) *TreeNode { * @param {number} key * @return {TreeNode} */ -var deleteNode = function (root, key) { - if (root === null) - return root; - if (root.val === key) { - if (!root.left) - return root.right; - else if (!root.right) - return root.left; - else { - let cur = root.right; - while (cur.left) { - cur = cur.left; - } - cur.left = root.left; - root = root.right; - delete root; - return root; - } - } - if (root.val > key) - root.left = deleteNode(root.left, key); - if (root.val < key) +var deleteNode = function(root, key) { + if (!root) return null; + if (key > root.val) { root.right = deleteNode(root.right, key); - return root; + return root; + } else if (key < root.val) { + root.left = deleteNode(root.left, key); + return root; + } else { + // 场景1: 该节点是叶节点 + if (!root.left && !root.right) { + return null + } + // 场景2: 有一个孩子节点不存在 + if (root.left && !root.right) { + return root.left; + } else if (root.right && !root.left) { + return root.right; + } + // 场景3: 左右节点都存在 + const rightNode = root.right; + // 获取最小值节点 + const minNode = getMinNode(rightNode); + // 将待删除节点的值替换为最小值节点值 + root.val = minNode.val; + // 删除最小值节点 + root.right = deleteNode(root.right, minNode.val); + return root; + } }; +function getMinNode(root) { + while (root.left) { + root = root.left; + } + return root; +} ``` 迭代 From 0f3ffb3e282af527cd2a529d7921a16321c2561e Mon Sep 17 00:00:00 2001 From: BaoTaoqi <464115280@qq.com> Date: Fri, 8 Jul 2022 10:58:53 +0800 Subject: [PATCH 106/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200746.=E4=BD=BF?= =?UTF-8?q?=E7=94=A8=E6=9C=80=E5=B0=8F=E8=8A=B1=E8=B4=B9=E7=88=AC=E6=A5=BC?= =?UTF-8?q?=E6=A2=AF.md=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0746.使用最小花费爬楼梯.md Rust版本 --- problems/0746.使用最小花费爬楼梯.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0746.使用最小花费爬楼梯.md b/problems/0746.使用最小花费爬楼梯.md index c2c73715..0006f7ac 100644 --- a/problems/0746.使用最小花费爬楼梯.md +++ b/problems/0746.使用最小花费爬楼梯.md @@ -288,6 +288,24 @@ function minCostClimbingStairs(cost: number[]): number { }; ``` +### Rust + +```Rust +use std::cmp::min; +impl Solution { + pub fn min_cost_climbing_stairs(cost: Vec) -> i32 { + let len = cost.len(); + let mut dp = vec![0; len]; + dp[0] = cost[0]; + dp[1] = cost[1]; + for i in 2..len { + dp[i] = min(dp[i-1], dp[i-2]) + cost[i]; + } + min(dp[len-1], dp[len-2]) + } +} +``` + ### C ```c From 9b3468e3b3e7ff844bc1c8d1a5bbb141072d5689 Mon Sep 17 00:00:00 2001 From: Ezralin <10881430+ezralin@user.noreply.gitee.com> Date: Fri, 8 Jul 2022 14:10:18 +0800 Subject: [PATCH 107/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A00034=20Kotlin?= =?UTF-8?q?=E7=89=88,0203=20Kotlin=E7=89=88,0209=20Koltin=E7=89=88,=200977?= =?UTF-8?q?=20Kotlin=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...排序数组中查找元素的第一个和最后一个位置.md | 43 +++++++++++++++++++ problems/0203.移除链表元素.md | 34 +++++++++++++++ problems/0209.长度最小的子数组.md | 32 ++++++++++++++ problems/0977.有序数组的平方.md | 28 ++++++++++++ 4 files changed, 137 insertions(+) diff --git a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md index a81b3641..6922c399 100644 --- a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md +++ b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md @@ -584,5 +584,48 @@ object Solution { ``` +### Kotlin +```kotlin +class Solution { + fun searchRange(nums: IntArray, target: Int): IntArray { + var index = binarySearch(nums, target) + // 没找到，返回[-1, -1] + if (index == -1) return intArrayOf(-1, -1) + var left = index + var right = index + // 寻找左边界 + while (left - 1 >=0 && nums[left - 1] == target){ + left-- + } + // 寻找右边界 + while (right + 1 target) { + right = middle - 1 + } + else { + if (nums[middle] < target) { + left = middle + 1 + } + else { + return middle + } + } + } + // 没找到，返回-1 + return -1 + } +} +``` + -----------------------
diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index b79d29a5..ea936093 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -532,5 +532,39 @@ object Solution { } } ``` +Kotlin: +```kotlin +/** + * Example: + * var li = ListNode(5) + * var v = li.`val` + * Definition for singly-linked list. + * class ListNode(var `val`: Int) { + * var next: ListNode? = null + * } + */ +class Solution { + fun removeElements(head: ListNode?, `val`: Int): ListNode? { + // 使用虚拟节点，令该节点指向head + var dummyNode = ListNode(-1) + dummyNode.next = head + // 使用cur遍历链表各个节点 + var cur = dummyNode + // 判断下个节点是否为空 + while (cur.next != null) { + // 符合条件，移除节点 + if (cur.next.`val` == `val`) { + cur.next = cur.next.next + } + // 不符合条件，遍历下一节点 + else { + cur = cur.next + } + } + // 注意：返回的不是虚拟节点 + return dummyNode.next + } +} +``` -----------------------
diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index 090e73f0..c1633def 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -417,6 +417,38 @@ class Solution { } } ``` +滑动窗口 +```kotlin +class Solution { + fun minSubArrayLen(target: Int, nums: IntArray): Int { + // 左边界 和 右边界 + var left: Int = 0 + var right: Int = 0 + // sum 用来记录和 + var sum: Int = 0 + // result记录一个固定值，便于判断是否存在的这样的数组 + var result: Int = Int.MAX_VALUE + // subLenth记录长度 + var subLength = Int.MAX_VALUE + + + while (right < nums.size) { + // 从数组首元素开始逐次求和 + sum += nums[right++] + // 判断 + while (sum >= target) { + var temp = right - left + // 每次和上一次比较求出最小数组长度 + subLength = if (subLength > temp) temp else subLength + // sum减少，左边界右移 + sum -= nums[left++] + } + } + // 如果subLength为初始值，则说明长度为0，否则返回subLength + return if(subLength == result) 0 else subLength + } +} +``` Scala: 滑动窗口: diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index 458107dd..d46b5259 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -362,6 +362,8 @@ class Solution { ``` Kotlin: + +双指针法 ```kotlin class Solution { // 双指针法 @@ -383,6 +385,32 @@ class Solution { } } ``` +骚操作（暴力思路） +```kotlin +class Solution { + fun sortedSquares(nums: IntArray): IntArray { + // left 与 right 用来控制循环，类似于滑动窗口 + var left: Int = 0; + var right: Int = nums.size - 1; + // 将每个数字的平方经过排序后加入result数值 + var result: IntArray = IntArray(nums.size); + var k: Int = nums.size - 1; + while (left <= right) { + // 从大到小，从后向前填满数组 + // [left, right] 控制循环 + if (nums[left] * nums[left] > nums[right] * nums[right]) { + result[k--] = nums[left] * nums[left] + left++ + } + else { + result[k--] = nums[right] * nums[right] + right-- + } + } + return result + } +} +``` Scala: From a8897483eade1262627d360544e7022951162243 Mon Sep 17 00:00:00 2001 From: BaoTaoqi <464115280@qq.com> Date: Fri, 8 Jul 2022 14:40:26 +0800 Subject: [PATCH 108/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200062.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E8=B7=AF=E5=BE=84.md=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0062.不同路径.md Rust版本 --- problems/0062.不同路径.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0062.不同路径.md b/problems/0062.不同路径.md index 02bcc2ae..aa7c8ab8 100644 --- a/problems/0062.不同路径.md +++ b/problems/0062.不同路径.md @@ -374,6 +374,30 @@ function uniquePaths(m: number, n: number): number { }; ``` +### Rust + +```Rust +impl Solution { + pub fn unique_paths(m: i32, n: i32) -> i32 { + let m = m as usize; + let n = n as usize; + let mut dp = vec![vec![0; n]; m]; + for i in 0..m { + dp[i][0] = 1; + } + for j in 0..n { + dp[0][j] = 1; + } + for i in 1..m { + for j in 1..n { + dp[i][j] = dp[i-1][j] + dp[i][j-1]; + } + } + dp[m-1][n-1] + } +} +``` + ### C ```c From 821cca116c6d42020ddac3bbb68c632eba7de132 Mon Sep 17 00:00:00 2001 From: BaoTaoqi <464115280@qq.com> Date: Fri, 8 Jul 2022 15:11:49 +0800 Subject: [PATCH 109/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200063.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E8=B7=AF=E5=BE=84II.md=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0063.不同路径II.md Rust版本 --- problems/0063.不同路径II.md | 36 ++++++++++++++++++++++++++++++++++++ 1 file changed, 36 insertions(+) diff --git a/problems/0063.不同路径II.md b/problems/0063.不同路径II.md index ca34055a..e3857db6 100644 --- a/problems/0063.不同路径II.md +++ b/problems/0063.不同路径II.md @@ -384,6 +384,42 @@ function uniquePathsWithObstacles(obstacleGrid: number[][]): number { }; ``` +### Rust + +```Rust +impl Solution { + pub fn unique_paths_with_obstacles(obstacle_grid: Vec>) -> i32 { + let m: usize = obstacle_grid.len(); + let n: usize = obstacle_grid[0].len(); + if obstacle_grid[0][0] == 1 || obstacle_grid[m-1][n-1] == 1 { + return 0; + } + let mut dp = vec![vec![0; n]; m]; + for i in 0..m { + if obstacle_grid[i][0] == 1 { + break; + } + else { dp[i][0] = 1; } + } + for j in 0..n { + if obstacle_grid[0][j] == 1 { + break; + } + else { dp[0][j] = 1; } + } + for i in 1..m { + for j in 1..n { + if obstacle_grid[i][j] == 1 { + continue; + } + dp[i][j] = dp[i-1][j] + dp[i][j-1]; + } + } + dp[m-1][n-1] + } +} +``` + ### C ```c From e2d16c5894b2b9abed99c42522977dac2b6e8654 Mon Sep 17 00:00:00 2001 From: BaoTaoqi <464115280@qq.com> Date: Sun, 10 Jul 2022 09:09:02 +0800 Subject: [PATCH 110/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200017.=E7=94=B5?= =?UTF-8?q?=E8=AF=9D=E5=8F=B7=E7=A0=81=E7=9A=84=E5=AD=97=E6=AF=8D=E7=BB=84?= =?UTF-8?q?=E5=90=88=200077.=E7=BB=84=E5=90=88=200077.=E7=BB=84=E5=90=88?= =?UTF-8?q?=E4=BC=98=E5=8C=96=200216.=E7=BB=84=E5=90=88=E6=80=BB=E5=92=8CI?= =?UTF-8?q?II=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0017.电话号码的字母组合 0077.组合 0077.组合优化 0216.组合总和III Rust版本 --- problems/0017.电话号码的字母组合.md | 43 +++++++++++++++++++++++++ problems/0077.组合.md | 50 +++++++++++++++++++++++++++++ problems/0077.组合优化.md | 26 +++++++++++++++ problems/0216.组合总和III.md | 29 +++++++++++++++++ 4 files changed, 148 insertions(+) diff --git a/problems/0017.电话号码的字母组合.md b/problems/0017.电话号码的字母组合.md index e357a33b..5778e903 100644 --- a/problems/0017.电话号码的字母组合.md +++ b/problems/0017.电话号码的字母组合.md @@ -454,6 +454,49 @@ function letterCombinations(digits: string): string[] { }; ``` +## Rust + +```Rust +impl Solution { + fn backtracking(result: &mut Vec, s: &mut String, map: &[&str; 10], digits: &String, index: usize) { + let len = digits.len(); + if len == index { + result.push(s.to_string()); + return; + } + // 在保证不会越界的情况下使用unwrap()将Some()中的值提取出来 + let digit= digits.chars().nth(index).unwrap().to_digit(10).unwrap() as usize; + let letters = map[digit]; + for i in letters.chars() { + s.push(i); + Self::backtracking(result, s, &map, &digits, index+1); + s.pop(); + } + } + pub fn letter_combinations(digits: String) -> Vec { + if digits.len() == 0 { + return vec![]; + } + const MAP: [&str; 10] = [ + "", + "", + "abc", + "def", + "ghi", + "jkl", + "mno", + "pqrs", + "tuv", + "wxyz" + ]; + let mut result: Vec = Vec::new(); + let mut s: String = String::new(); + Self::backtracking(&mut result, &mut s, &MAP, &digits, 0); + result + } +} +``` + ## C ```c diff --git a/problems/0077.组合.md b/problems/0077.组合.md index fc72be15..5a4811ba 100644 --- a/problems/0077.组合.md +++ b/problems/0077.组合.md @@ -535,6 +535,56 @@ func backtrack(n,k,start int,track []int){ } ``` +### Rust + +```Rust +impl Solution { + fn backtracking(result: &mut Vec>, path: &mut Vec, n: i32, k: i32, startIndex: i32) { + let len= path.len() as i32; + if len == k{ + result.push(path.to_vec()); + return; + } + for i in startIndex..= n { + path.push(i); + Self::backtracking(result, path, n, k, i+1); + path.pop(); + } + } + pub fn combine(n: i32, k: i32) -> Vec> { + let mut result: Vec> = Vec::new(); + let mut path: Vec = Vec::new(); + Self::backtracking(&mut result, &mut path, n, k, 1); + result + } +} +``` + +剪枝 +```Rust +impl Solution { + fn backtracking(result: &mut Vec>, path: &mut Vec, n: i32, k: i32, startIndex: i32) { + let len= path.len() as i32; + if len == k{ + result.push(path.to_vec()); + return; + } + // 此处剪枝 + for i in startIndex..= n - (k - len) + 1 { + path.push(i); + Self::backtracking(result, path, n, k, i+1); + path.pop(); + } + } + pub fn combine(n: i32, k: i32) -> Vec> { + let mut result: Vec> = Vec::new(); + let mut path: Vec = Vec::new(); + Self::backtracking(&mut result, &mut path, n, k, 1); + result + } +} +``` + ### C ```c int* path; diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md index 8d742566..e336fb75 100644 --- a/problems/0077.组合优化.md +++ b/problems/0077.组合优化.md @@ -261,6 +261,32 @@ function combine(n: number, k: number): number[][] { }; ``` +Rust: + +```Rust +impl Solution { + fn backtracking(result: &mut Vec>, path: &mut Vec, n: i32, k: i32, startIndex: i32) { + let len= path.len() as i32; + if len == k{ + result.push(path.to_vec()); + return; + } + // 此处剪枝 + for i in startIndex..= n - (k - len) + 1 { + path.push(i); + Self::backtracking(result, path, n, k, i+1); + path.pop(); + } + } + pub fn combine(n: i32, k: i32) -> Vec> { + let mut result: Vec> = Vec::new(); + let mut path: Vec = Vec::new(); + Self::backtracking(&mut result, &mut path, n, k, 1); + result + } +} +``` + C: ```c diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md index 2c1bf717..0a59e216 100644 --- a/problems/0216.组合总和III.md +++ b/problems/0216.组合总和III.md @@ -411,6 +411,35 @@ function combinationSum3(k: number, n: number): number[][] { }; ``` +## Rust + +```Rust +impl Solution { + fn backtracking(result: &mut Vec>, path:&mut Vec, targetSum:i32, k: i32, mut sum: i32, startIndex: i32) { + let len = path.len() as i32; + if len == k { + if sum == targetSum { + result.push(path.to_vec()); + } + return; + } + for i in startIndex..=9 { + sum += i; + path.push(i); + Self::backtracking(result, path, targetSum, k, sum, i+1); + sum -= i; + path.pop(); + } + } + pub fn combination_sum3(k: i32, n: i32) -> Vec> { + let mut result: Vec> = Vec::new(); + let mut path: Vec = Vec::new(); + Self::backtracking(&mut result, &mut path, n, k, 0, 1); + result + } +} +``` + ## C ```c From 85e0d6c85d593339152483a3fea33ab23cdcd5b3 Mon Sep 17 00:00:00 2001 From: AronJudge <2286381138@qq.com> Date: Mon, 11 Jul 2022 11:06:51 +0800 Subject: [PATCH 111/136] =?UTF-8?q?=E6=9B=B4=E6=96=B00347=20=E5=89=8DK?= =?UTF-8?q?=E4=B8=AA=E9=AB=98=E9=A2=91=E5=85=83=E7=B4=A0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 由原来的构建小顶堆改为构建大顶堆，减少部分代码逻辑，并增加了注释 通过LeetCode提交代码，修改后的执行时间更快。 --- problems/0347.前K个高频元素.md | 7 ++----- 1 file changed, 2 insertions(+), 5 deletions(-) diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md index a04041cc..0c978b2a 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -141,13 +141,10 @@ class Solution { } Set> entries = map.entrySet(); - // 根据map的value值正序排，相当于一个小顶堆 - PriorityQueue> queue = new PriorityQueue<>((o1, o2) -> o1.getValue() - o2.getValue()); + // 根据map的value值，构建于一个大顶堆（o1 - o2: 小顶堆， o2 - o1 : 大顶堆） + PriorityQueue> queue = new PriorityQueue<>((o1, o2) -> o2.getValue() - o1.getValue()); for (Map.Entry entry : entries) { queue.offer(entry); - if (queue.size() > k) { - queue.poll(); - } } for (int i = k - 1; i >= 0; i--) { result[i] = queue.poll().getKey(); From c265c9220c62a78109c1f6033d47f9bc8b3f47bd Mon Sep 17 00:00:00 2001 From: BaoTaoqi <464115280@qq.com> Date: Mon, 11 Jul 2022 14:50:03 +0800 Subject: [PATCH 112/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200203.=E7=A7=BB?= =?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0=200344.=E5=8F=8D?= =?UTF-8?q?=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2=200376.=E6=91=86=E5=8A=A8?= =?UTF-8?q?=E5=BA=8F=E5=88=97=200541.=E5=8F=8D=E8=BD=AC=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2II=20=E5=89=91=E6=8C=87Offer05.=E6=9B=BF=E6=8D=A2?= =?UTF-8?q?=E7=A9=BA=E6=A0=BC=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0203.移除链表元素 0344.反转字符串 0376.摆动序列 0541.反转字符串II 剑指Offer05.替换空格 Rust版本 --- problems/0203.移除链表元素.md | 18 ++++++++++-------- problems/0344.反转字符串.md | 16 ++++++++++++++++ problems/0376.摆动序列.md | 23 +++++++++++++++++++++++ problems/0541.反转字符串II.md | 31 +++++++++++++++++++++++++++++++ problems/剑指Offer05.替换空格.md | 31 +++++++++++++++++++++++++++++++ 5 files changed, 111 insertions(+), 8 deletions(-) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index b79d29a5..cb46dc70 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -487,17 +487,19 @@ RUST: // } impl Solution { pub fn remove_elements(head: Option>, val: i32) -> Option> { - let mut head = head; - let mut dummy_head = ListNode::new(0); - let mut cur = &mut dummy_head; - while let Some(mut node) = head { - head = std::mem::replace(&mut node.next, None); - if node.val != val { - cur.next = Some(node); + let mut dummyHead = Box::new(ListNode::new(0)); + dummyHead.next = head; + let mut cur = dummyHead.as_mut(); + // 使用take()替换std::men::replace(&mut node.next, None)达到相同的效果，并且更普遍易读 + while let Some(nxt) = cur.next.take() { + if nxt.val == val { + cur.next = nxt.next; + } else { + cur.next = Some(nxt); cur = cur.next.as_mut().unwrap(); } } - dummy_head.next + dummyHead.next } } ``` diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index aba6e2f3..3138f60f 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -238,6 +238,22 @@ func reverseString(_ s: inout [Character]) { ``` +Rust: +```Rust +impl Solution { + pub fn reverse_string(s: &mut Vec) { + let (mut left, mut right) = (0, s.len()-1); + while left < right { + let temp = s[left]; + s[left] = s[right]; + s[right] = temp; + left += 1; + right -= 1; + } + } +} +``` + C: ```c void reverseString(char* s, int sSize){ diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index 00f8f70c..33a5282d 100644 --- a/problems/0376.摆动序列.md +++ b/problems/0376.摆动序列.md @@ -298,6 +298,29 @@ var wiggleMaxLength = function(nums) { }; ``` +### Rust +**贪心** +```Rust +impl Solution { + pub fn wiggle_max_length(nums: Vec) -> i32 { + let len = nums.len() as usize; + if len <= 1 { + return len as i32; + } + let mut preDiff = 0; + let mut curDiff = 0; + let mut result = 1; + for i in 0..len-1 { + curDiff = nums[i+1] - nums[i]; + if (preDiff <= 0 && curDiff > 0) || (preDiff >= 0 && curDiff < 0) { + result += 1; + preDiff = curDiff; + } + } + result + } +} +``` ### C **贪心** diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md index 84061ef5..7ef6463e 100644 --- a/problems/0541.反转字符串II.md +++ b/problems/0541.反转字符串II.md @@ -389,5 +389,36 @@ object Solution { } } ``` + +Rust: + +```Rust +impl Solution { + pub fn reverse(s: &mut Vec, mut begin: usize, mut end: usize){ + while begin < end { + let temp = s[begin]; + s[begin] = s[end]; + s[end] = temp; + begin += 1; + end -= 1; + } + } + pub fn reverse_str(s: String, k: i32) -> String { + let len = s.len(); + let k = k as usize; + let mut s = s.chars().collect::>(); + for i in (0..len).step_by(2 * k) { + if i + k < len { + Self::reverse(&mut s, i, i + k - 1); + } + else { + Self::reverse(&mut s, i, len - 1); + } + } + s.iter().collect::() + } +} +``` + -----------------------
diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index 78b03b17..e1ccc458 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -506,6 +506,37 @@ function spaceLen($s){ } ``` +Rust + +```Rust +impl Solution { + pub fn replace_space(s: String) -> String { + let mut len: usize = s.len(); + let mut s = s.chars().collect::>(); + let mut count = 0; + for i in &s { + if i.is_ascii_whitespace() { + count += 1; + } + } + let mut new_len = len + count * 2; + s.resize(new_len, ' '); + while len < new_len { + len -= 1; + new_len -= 1; + if s[len].is_ascii_whitespace() { + s[new_len] = '0'; + s[new_len - 1] = '2'; + s[new_len - 2] = '%'; + new_len -= 2; + } + else { s[new_len] = s[len] } + } + s.iter().collect::() + } +} +``` + -----------------------
From e0a422859904ddae865cc828bc30d6620ccb8137 Mon Sep 17 00:00:00 2001 From: Ezralin <10881430+ezralin@user.noreply.gitee.com> Date: Mon, 11 Jul 2022 15:09:41 +0800 Subject: [PATCH 113/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A00206.=E7=BF=BB?= =?UTF-8?q?=E8=BD=AC=E9=93=BE=E8=A1=A8=20Kotlin=E8=A7=A3=E6=9E=90=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0206.翻转链表.md | 35 +++++++++++++++++++++++++++++++++++ 1 file changed, 35 insertions(+) diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index 24ec7b94..bb9b62d5 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -420,6 +420,41 @@ fun reverseList(head: ListNode?): ListNode? { return pre } ``` +```kotlin +/** + * Example: + * var li = ListNode(5) + * var v = li.`val` + * Definition for singly-linked list. + * class ListNode(var `val`: Int) { + * var next: ListNode? = null + * } + */ +class Solution { + fun reverseList(head: ListNode?): ListNode? { + // temp用来存储临时的节点 + var temp: ListNode? + // cur用来遍历链表 + var cur: ListNode? = head + // pre用来作为链表反转的工具 + // pre是比pre前一位的节点 + var pre: ListNode? = null + while (cur != null) { + // 临时存储原本cur的下一个节点 + temp = cur.next + // 使cur下一节点地址为它之前的 + cur.next = pre + // 之后随着cur的遍历移动pre + pre = cur; + // 移动cur遍历链表各个节点 + cur = temp; + } + // 由于开头使用pre为null,所以cur等于链表本身长度+1， + // 此时pre在cur前一位，所以此时pre为头节点 + return pre; + } +} +``` Swift： ```swift From 4e77e911b397938419a41c2a06684318740fc907 Mon Sep 17 00:00:00 2001 From: cezarbbb <105843128+cezarbbb@users.noreply.github.com> Date: Mon, 11 Jul 2022 17:42:21 +0800 Subject: [PATCH 114/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200151.=E7=BF=BB?= =?UTF-8?q?=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95?= =?UTF-8?q?=E8=AF=8D=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0151.翻转字符串里的单词 Rust版本 --- problems/0151.翻转字符串里的单词.md | 51 +++++++++++++++++++++++++++++ 1 file changed, 51 insertions(+) diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index 38372f91..552584f5 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -864,7 +864,58 @@ function reverseString(&$s, $start, $end) { return ; } ``` +Rust: +```Rust +// 根据C++版本二思路进行实现 +// 函数名根据Rust编译器建议由驼峰命名法改为蛇形命名法 +impl Solution { + pub fn reverse(s: &mut Vec, mut begin: usize, mut end: usize){ + while begin < end { + let temp = s[begin]; + s[begin] = s[end]; + s[end] = temp; + begin += 1; + end -= 1; + } +} +pub fn remove_extra_spaces(s: &mut Vec) { + let mut slow: usize = 0; + let len = s.len(); + // 注意这里不能用for循环，不然在里面那个while循环中对i的递增会失效 + let mut i: usize = 0; + while i < len { + if !s[i].is_ascii_whitespace() { + if slow != 0 { + s[slow] = ' '; + slow += 1; + } + while i < len && !s[i].is_ascii_whitespace() { + s[slow] = s[i]; + slow += 1; + i += 1; + } + } + i += 1; + } + s.resize(slow, ' '); + } + pub fn reverse_words(s: String) -> String { + let mut s = s.chars().collect::>(); + Self::remove_extra_spaces(&mut s); + let len = s.len(); + Self::reverse(&mut s, 0, len - 1); + let mut start = 0; + for i in 0..=len { + if i == len || s[i].is_ascii_whitespace() { + Self::reverse(&mut s, start, i - 1); + start = i + 1; + } + } + s.iter().collect::() + } +} +``` -----------------------
From ba081f84a85ff75134c4db551e0e631b86693081 Mon Sep 17 00:00:00 2001 From: AronJudge <2286381138@qq.com> Date: Tue, 12 Jul 2022 11:36:26 +0800 Subject: [PATCH 115/136] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200104.=20=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E6=9C=80=E5=A4=A7=E6=B7=B1=E5=BA=A6?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 代码错误，更正代码，通过LeetCode代码检验。 --- problems/0104.二叉树的最大深度.md | 25 ++++++++++++------------- 1 file changed, 12 insertions(+), 13 deletions(-) diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md index ed27f95d..55980189 100644 --- a/problems/0104.二叉树的最大深度.md +++ b/problems/0104.二叉树的最大深度.md @@ -294,14 +294,13 @@ class solution { /** * 递归法 */ - public int maxdepth(treenode root) { + public int maxDepth(TreeNode root) { if (root == null) { return 0; } - int leftdepth = maxdepth(root.left); - int rightdepth = maxdepth(root.right); - return math.max(leftdepth, rightdepth) + 1; - + int leftDepth = maxDepth(root.left); + int rightDepth = maxDepth(root.right); + return Math.max(leftDepth, rightDepth) + 1; } } ``` @@ -311,23 +310,23 @@ class solution { /** * 迭代法，使用层序遍历 */ - public int maxdepth(treenode root) { + public int maxDepth(TreeNode root) { if(root == null) { return 0; } - deque deque = new linkedlist<>(); + Deque deque = new LinkedList<>(); deque.offer(root); int depth = 0; - while (!deque.isempty()) { + while (!deque.isEmpty()) { int size = deque.size(); depth++; for (int i = 0; i < size; i++) { - treenode poll = deque.poll(); - if (poll.left != null) { - deque.offer(poll.left); + TreeNode node = deque.poll(); + if (node.left != null) { + deque.offer(node.left); } - if (poll.right != null) { - deque.offer(poll.right); + if (node.right != null) { + deque.offer(node.right); } } } From a8e19d60bb371a44296134932b7470825c1fda6a Mon Sep 17 00:00:00 2001 From: cezarbbb <105843128+cezarbbb@users.noreply.github.com> Date: Tue, 12 Jul 2022 14:08:50 +0800 Subject: [PATCH 116/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200202.=E5=BF=AB?= =?UTF-8?q?=E4=B9=90=E6=95=B0=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0202.快乐数 Rust版本 --- problems/0202.快乐数.md | 30 ++++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 0bea0c72..2303765c 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -315,6 +315,36 @@ class Solution { } ``` +Rust: +```Rust +use std::collections::HashSet; +impl Solution { + pub fn get_sum(mut n: i32) -> i32 { + let mut sum = 0; + while n > 0 { + sum += (n % 10) * (n % 10); + n /= 10; + } + sum + } + + pub fn is_happy(n: i32) -> bool { + let mut n = n; + let mut set = HashSet::new(); + loop { + let sum = Self::get_sum(n); + if sum == 1 { + return true; + } + if set.contains(&sum) { + return false; + } else { set.insert(sum); } + n = sum; + } + } +} +``` + C: ```C typedef struct HashNodeTag { From eb3eb336dedfcbbcc0d90eab392e4799479d675e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E6=A1=9C=E5=B0=8F=E8=B7=AF=E4=B8=83=E8=91=89?= <20304773@qq.com> Date: Tue, 12 Jul 2022 14:09:05 +0800 Subject: [PATCH 117/136] =?UTF-8?q?Update=200059.=E8=9E=BA=E6=97=8B?= =?UTF-8?q?=E7=9F=A9=E9=98=B5II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0059.螺旋矩阵II.md | 25 +++++++++++++++++++++++++ 1 file changed, 25 insertions(+) diff --git a/problems/0059.螺旋矩阵II.md b/problems/0059.螺旋矩阵II.md index bf0a279e..9690abb6 100644 --- a/problems/0059.螺旋矩阵II.md +++ b/problems/0059.螺旋矩阵II.md @@ -598,5 +598,30 @@ object Solution { } } ``` +C#： +```csharp +public class Solution { + public int[][] GenerateMatrix(int n) { + int[][] answer = new int[n][]; + for(int i = 0; i < n; i++) + answer[i] = new int[n]; + int start = 0; + int end = n - 1; + int tmp = 1; + while(tmp < n * n) + { + for(int i = start; i < end; i++) answer[start][i] = tmp++; + for(int i = start; i < end; i++) answer[i][end] = tmp++; + for(int i = end; i > start; i--) answer[end][i] = tmp++; + for(int i = end; i > start; i--) answer[i][start] = tmp++; + start++; + end--; + } + if(n % 2 == 1) answer[n / 2][n / 2] = tmp; + return answer; + } +} +``` + -----------------------
From 07ab44ede3fa043184754a638f59e6b76a7c3e0a Mon Sep 17 00:00:00 2001 From: cezarbbb <105843128+cezarbbb@users.noreply.github.com> Date: Tue, 12 Jul 2022 15:26:07 +0800 Subject: [PATCH 118/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200015.=E4=B8=89?= =?UTF-8?q?=E6=95=B0=E4=B9=8B=E5=92=8C=200018.=E5=9B=9B=E6=95=B0=E4=B9=8B?= =?UTF-8?q?=E5=92=8C=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0015.三数之和 0018.四数之和 Rust版本 --- problems/0015.三数之和.md | 65 +++++++++++++++++++++++++++++++++++++++ problems/0018.四数之和.md | 45 +++++++++++++++++++++++++++ 2 files changed, 110 insertions(+) diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md index 4f1d711a..94885b0c 100644 --- a/problems/0015.三数之和.md +++ b/problems/0015.三数之和.md @@ -554,6 +554,71 @@ func threeSum(_ nums: [Int]) -> [[Int]] { } ``` +Rust: +```Rust +// 哈希解法 +use std::collections::HashSet; +impl Solution { + pub fn three_sum(nums: Vec) -> Vec> { + let mut result: Vec> = Vec::new(); + let mut nums = nums; + nums.sort(); + let len = nums.len(); + for i in 0..len { + if nums[i] > 0 { break; } + if i > 0 && nums[i] == nums[i - 1] { continue; } + let mut set = HashSet::new(); + for j in (i + 1)..len { + if j > i + 2 && nums[j] == nums[j - 1] && nums[j] == nums[j - 2] { continue; } + let c = 0 - (nums[i] + nums[j]); + if set.contains(&c) { + result.push(vec![nums[i], nums[j], c]); + set.remove(&c); + } else { set.insert(nums[j]); } + } + } + result + } +} +``` + +```Rust +// 双指针法 +use std::collections::HashSet; +impl Solution { + pub fn three_sum(nums: Vec) -> Vec> { + let mut result: Vec> = Vec::new(); + let mut nums = nums; + nums.sort(); + let len = nums.len(); + for i in 0..len { + if nums[i] > 0 { return result; } + if i > 0 && nums[i] == nums[i - 1] { continue; } + let (mut left, mut right) = (i + 1, len - 1); + while left < right { + if nums[i] + nums[left] + nums[right] > 0 { + right -= 1; + // 去重 + while left < right && nums[right] == nums[right + 1] { right -= 1; } + } else if nums[i] + nums[left] + nums[right] < 0 { + left += 1; + // 去重 + while left < right && nums[left] == nums[left - 1] { left += 1; } + } else { + result.push(vec![nums[i], nums[left], nums[right]]); + // 去重 + right -= 1; + left += 1; + while left < right && nums[right] == nums[right + 1] { right -= 1; } + while left < right && nums[left] == nums[left - 1] { left += 1; } + } + } + } + result + } +} +``` + C: ```C //qsort辅助cmp函数 diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 2146a114..f6cfad29 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -522,6 +522,51 @@ public class Solution } } ``` + +Rust: +```Rust +impl Solution { + pub fn four_sum(nums: Vec, target: i32) -> Vec> { + let mut result: Vec> = Vec::new(); + let mut nums = nums; + nums.sort(); + let len = nums.len(); + for k in 0..len { + // 剪枝 + if nums[k] > target && (nums[k] > 0 || target > 0) { break; } + // 去重 + if k > 0 && nums[k] == nums[k - 1] { continue; } + for i in (k + 1)..len { + // 剪枝 + if nums[k] + nums[i] > target && (nums[k] + nums[i] >= 0 || target >= 0) { break; } + // 去重 + if i > k + 1 && nums[i] == nums[i - 1] { continue; } + let (mut left, mut right) = (i + 1, len - 1); + while left < right { + if nums[k] + nums[i] > target - (nums[left] + nums[right]) { + right -= 1; + // 去重 + while left < right && nums[right] == nums[right + 1] { right -= 1; } + } else if nums[k] + nums[i] < target - (nums[left] + nums[right]) { + left += 1; + // 去重 + while left < right && nums[left] == nums[left - 1] { left += 1; } + } else { + result.push(vec![nums[k], nums[i], nums[left], nums[right]]); + // 去重 + while left < right && nums[right] == nums[right - 1] { right -= 1; } + while left < right && nums[left] == nums[left + 1] { left += 1; } + left += 1; + right -= 1; + } + } + } + } + result + } +} +``` + Scala: ```scala object Solution { From f5e1834439d64d09b2e8d8b4e5141510e205dbc3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E6=9D=A8=E6=9D=A8?= <2979742951@qq.com> Date: Wed, 13 Jul 2022 16:15:12 +0800 Subject: [PATCH 119/136] =?UTF-8?q?=E5=8F=8D=E8=BD=AC=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2II=E6=B7=BB=E5=8A=A0=E4=BA=86=E4=B8=80=E7=A7=8DC++?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0541.反转字符串II.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md index 84061ef5..d9b9466c 100644 --- a/problems/0541.反转字符串II.md +++ b/problems/0541.反转字符串II.md @@ -63,6 +63,24 @@ public: }; ``` + +``` +class Solution { +public: + string reverseStr(string s, int k) { + int n=s.size(),pos=0; + while(pos Date: Wed, 13 Jul 2022 19:27:34 +0800 Subject: [PATCH 120/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E5=89=91=E6=8C=87?= =?UTF-8?q?Offer=2058-II.=E5=B7=A6=E6=97=8B=E8=BD=AC=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 剑指Offer 58-II.左旋转字符串 Rust版本 --- problems/剑指Offer58-II.左旋转字符串.md | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/problems/剑指Offer58-II.左旋转字符串.md b/problems/剑指Offer58-II.左旋转字符串.md index 4674c141..de4a9030 100644 --- a/problems/剑指Offer58-II.左旋转字符串.md +++ b/problems/剑指Offer58-II.左旋转字符串.md @@ -341,7 +341,30 @@ object Solution { } ``` +Rust: +```Rust +impl Solution { + pub fn reverse(s: &mut Vec, mut begin: usize, mut end: usize){ + while begin < end { + let temp = s[begin]; + s[begin] = s[end]; + s[end] = temp; + begin += 1; + end -= 1; + } + } + pub fn reverse_left_words(s: String, n: i32) -> String { + let len = s.len(); + let mut s = s.chars().collect::>(); + let n = n as usize; + Self::reverse(&mut s, 0, n - 1); + Self::reverse(&mut s, n, len - 1); + Self::reverse(&mut s, 0, len - 1); + s.iter().collect::() + } +} +``` From acd71b8b9e134fdc1fbe0f73d78b5f80b7b03a59 Mon Sep 17 00:00:00 2001 From: cezarbbb <105843128+cezarbbb@users.noreply.github.com> Date: Thu, 14 Jul 2022 15:46:41 +0800 Subject: [PATCH 121/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200028.=E5=AE=9E?= =?UTF-8?q?=E7=8E=B0strStr=200459.=E9=87=8D=E5=A4=8D=E7=9A=84=E5=AD=90?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0028.实现strStr 0459.重复的子字符串 Rust版本 --- problems/0028.实现strStr.md | 44 +++++++++++++++++++++++++++++++++ problems/0459.重复的子字符串.md | 32 ++++++++++++++++++++++++ 2 files changed, 76 insertions(+) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index a95a52f8..1887e91b 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -1241,5 +1241,49 @@ function getNext(&$next, $s){ } } ``` + +Rust: + +> 前缀表统一不减一 +```Rust +impl Solution { + pub fn get_next(next: &mut Vec, s: &Vec) { + let len = s.len(); + let mut j = 0; + for i in 1..len { + while j > 0 && s[i] != s[j] { + j = next[j - 1]; + } + if s[i] == s[j] { + j += 1; + } + next[i] = j; + } + } + + pub fn str_str(haystack: String, needle: String) -> i32 { + let (haystack_len, needle_len) = (haystack.len(), needle.len()); + if haystack_len == 0 { return 0; } + if haystack_len < needle_len { return -1;} + let (haystack, needle) = (haystack.chars().collect::>(), needle.chars().collect::>()); + let mut next: Vec = vec![0; haystack_len]; + Self::get_next(&mut next, &needle); + let mut j = 0; + for i in 0..haystack_len { + while j > 0 && haystack[i] != needle[j] { + j = next[j - 1]; + } + if haystack[i] == needle[j] { + j += 1; + } + if j == needle_len { + return (i - needle_len + 1) as i32; + } + } + return -1; + } +} +``` + -----------------------
diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index bb55bd7c..4f45f4d7 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -505,5 +505,37 @@ Swift: } ``` +Rust: + +>前缀表统一不减一 +```Rust +impl Solution { + pub fn get_next(next: &mut Vec, s: &Vec) { + let len = s.len(); + let mut j = 0; + for i in 1..len { + while j > 0 && s[i] != s[j] { + j = next[j - 1]; + } + if s[i] == s[j] { + j += 1; + } + next[i] = j; + } + } + + pub fn repeated_substring_pattern(s: String) -> bool { + let s = s.chars().collect::>(); + let len = s.len(); + if len == 0 { return false; }; + let mut next = vec![0; len]; + Self::get_next(&mut next, &s); + if next[len - 1] != 0 && len % (len - (next[len - 1] )) == 0 { return true; } + return false; + } +} +``` + + -----------------------
From 781076a73d7d9de9f0fe525cbc8404358ca59990 Mon Sep 17 00:00:00 2001 From: Tristone Date: Thu, 14 Jul 2022 22:39:17 +0800 Subject: [PATCH 122/136] =?UTF-8?q?=E9=9C=80=E8=A6=81=E5=90=8C=E6=97=B6?= =?UTF-8?q?=E4=BF=AE=E6=94=B9=E8=BE=93=E5=85=A5=E6=95=B0=E7=BB=84nums?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0027.移除元素.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 9d687cfb..fd24e8f8 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -219,6 +219,7 @@ func removeElement(nums []int, val int) int { res++ } } + nums=nums[:res] return res } ``` From 26f5e59cdaf5536e0b08b94d554e4ef4428732a7 Mon Sep 17 00:00:00 2001 From: leo <55868230+LIU-HONGYANG@users.noreply.github.com> Date: Sat, 16 Jul 2022 11:56:18 +0800 Subject: [PATCH 123/136] =?UTF-8?q?Update=20=E9=9D=A2=E8=AF=95=E9=A2=9802.?= =?UTF-8?q?07.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/面试题02.07.链表相交.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index f603925d..d799bc80 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -208,7 +208,7 @@ func getIntersectionNode(headA, headB *ListNode) *ListNode { } ``` -递归版本： +双指针 ```go func getIntersectionNode(headA, headB *ListNode) *ListNode { From fb982fa79d643bb3e01960bb01897322cc424bf7 Mon Sep 17 00:00:00 2001 From: xiaojun <13589818805@163.com> Date: Sat, 16 Jul 2022 15:00:22 +0800 Subject: [PATCH 124/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=88234.?= =?UTF-8?q?=E5=9B=9E=E6=96=87=E9=93=BE=E8=A1=A8=EF=BC=89=E7=9A=84go?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0234.回文链表.md | 68 +++++++++++++++++++++++++++++++++++++++ 1 file changed, 68 insertions(+) diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md index eeee6fa5..1ac8756a 100644 --- a/problems/0234.回文链表.md +++ b/problems/0234.回文链表.md @@ -276,7 +276,75 @@ class Solution: ### Go ```go +/** + * Definition for singly-linked list. + * type ListNode struct { + * Val int + * Next *ListNode + * } + */ +//方法一，使用数组 +func isPalindrome(head *ListNode) bool{ + //计算切片长度，避免切片频繁扩容 + cur,ln:=head,0 + for cur!=nil{ + ln++ + cur=cur.Next + } + nums:=make([]int,ln) + index:=0 + for head!=nil{ + nums[index]=head.Val + index++ + head=head.Next + } + //比较回文切片 + for i,j:=0,ln-1;i<=j;i,j=i+1,j-1{ + if nums[i]!=nums[j]{return false} + } + return true +} +// 方法二，快慢指针 +func isPalindrome(head *ListNode) bool { + if head==nil&&head.Next==nil{return true} + //慢指针，找到链表中间分位置，作为分割 + slow:=head + fast:=head + //记录慢指针的前一个节点，用来分割链表 + pre:=head + for fast!=nil && fast.Next!=nil{ + pre=slow + slow=slow.Next + fast=fast.Next.Next + } + //分割链表 + pre.Next=nil + //前半部分 + cur1:=head + //反转后半部分，总链表长度如果是奇数，cur2比cur1多一个节点 + cur2:=ReverseList(slow) + + //开始两个链表的比较 + for cur1!=nil{ + if cur1.Val!=cur2.Val{return false} + cur1=cur1.Next + cur2=cur2.Next + } + return true +} +//反转链表 +func ReverseList(head *ListNode) *ListNode{ + var pre *ListNode + cur:=head + for cur!=nil{ + tmp:=cur.Next + cur.Next=pre + pre=cur + cur=tmp + } + return pre +} ``` ### JavaScript From 8b9b64d7d576a783a3bb8fecdc3d614a4f9a6f36 Mon Sep 17 00:00:00 2001 From: AronJudge <2286381138@qq.com> Date: Sat, 16 Jul 2022 15:43:50 +0800 Subject: [PATCH 125/136] =?UTF-8?q?=E4=BF=AE=E6=94=B9:=200617=20=E5=90=88?= =?UTF-8?q?=E5=B9=B6=E4=BA=8C=E5=8F=89=E6=A0=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改代码细节，优化性能. 用原有的Root1 代替 创建的NewRoot. --- problems/0617.合并二叉树.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md index acdcc0aa..6a843763 100644 --- a/problems/0617.合并二叉树.md +++ b/problems/0617.合并二叉树.md @@ -262,10 +262,10 @@ class Solution { if (root1 == null) return root2; if (root2 == null) return root1; - TreeNode newRoot = new TreeNode(root1.val + root2.val); - newRoot.left = mergeTrees(root1.left,root2.left); - newRoot.right = mergeTrees(root1.right,root2.right); - return newRoot; + root1.val += root2.val; + root1.left = mergeTrees(root1.left,root2.left); + root1.right = mergeTrees(root1.right,root2.right); + return root1; } } ``` From 6f3dc1e6f80c5527d266c125b1a57366d314787a Mon Sep 17 00:00:00 2001 From: DarrenRuan <31378679+DarrenRuan@users.noreply.github.com> Date: Sat, 16 Jul 2022 23:57:33 -0400 Subject: [PATCH 126/136] =?UTF-8?q?Update=200039.=E7=BB=84=E5=90=88?= =?UTF-8?q?=E6=80=BB=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 这里python代码有个注释似乎写错了。原来为`不是i-1`，应该是`不是i+1`. --- problems/0039.组合总和.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0039.组合总和.md b/problems/0039.组合总和.md index 564d13ea..54e9f2e5 100644 --- a/problems/0039.组合总和.md +++ b/problems/0039.组合总和.md @@ -291,7 +291,7 @@ class Solution: for i in range(start_index, len(candidates)): sum_ += candidates[i] self.path.append(candidates[i]) - self.backtracking(candidates, target, sum_, i) # 因为无限制重复选取，所以不是i-1 + self.backtracking(candidates, target, sum_, i) # 因为无限制重复选取，所以不是i+1 sum_ -= candidates[i] # 回溯 self.path.pop() # 回溯 ``` From 50b9753b647ace27caa0a1a31bd91975672db4d7 Mon Sep 17 00:00:00 2001 From: dcj_hp <294487055@qq.com> Date: Wed, 20 Jul 2022 01:11:00 +0800 Subject: [PATCH 127/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200376.=E6=91=86?= =?UTF-8?q?=E5=8A=A8=E5=BA=8F=E5=88=97=20=20=E5=8A=A8=E6=80=81=E8=A7=84?= =?UTF-8?q?=E5=88=92=E7=89=88=E6=9C=AC(Python,=20C)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0376.摆动序列.md | 64 +++++++++++++++++++++++++++++++++++++++ 1 file changed, 64 insertions(+) diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index 70a62145..714dba57 100644 --- a/problems/0376.摆动序列.md +++ b/problems/0376.摆动序列.md @@ -228,6 +228,8 @@ class Solution { ### Python +**贪心** + ```python class Solution: def wiggleMaxLength(self, nums: List[int]) -> int: @@ -240,7 +242,30 @@ class Solution: return res ``` +**动态规划** + +```python +class Solution: + def wiggleMaxLength(self, nums: List[int]) -> int: + # 0 i 作为波峰的最大长度 + # 1 i 作为波谷的最大长度 + # dp是一个列表，列表中每个元素是长度为 2 的列表 + dp = [] + for i in range(len(nums)): + # 初始为[1, 1] + dp.append([1, 1]) + for j in range(i): + # nums[i] 为波谷 + if nums[j] > nums[i]: + dp[i][1] = max(dp[i][1], dp[j][0] + 1) + # nums[i] 为波峰 + if nums[j] < nums[i]: + dp[i][0] = max(dp[i][0], dp[j][1] + 1) + return max(dp[-1][0], dp[-1][1]) +``` + ### Go + ```golang func wiggleMaxLength(nums []int) int { var count,preDiff,curDiff int @@ -324,6 +349,7 @@ impl Solution { ### C **贪心** + ```c int wiggleMaxLength(int* nums, int numsSize){ if(numsSize <= 1) @@ -349,6 +375,44 @@ int wiggleMaxLength(int* nums, int numsSize){ } ``` +**动态规划** + +```c +int max(int left, int right) +{ + return left > right ? left : right; +} +int wiggleMaxLength(int* nums, int numsSize){ + if(numsSize <= 1) + { + return numsSize; + } + // 0 i 作为波峰的最大长度 + // 1 i 作为波谷的最大长度 + int dp[numsSize][2]; + for(int i = 0; i < numsSize; i++) + { + dp[i][0] = 1; + dp[i][1] = 1; + for(int j = 0; j < i; j++) + { + // nums[i] 为山谷 + if(nums[j] > nums[i]) + { + dp[i][1] = max(dp[i][1], dp[j][0] + 1); + } + // nums[i] 为山峰 + if(nums[j] < nums[i]) + { + dp[i][0] = max(dp[i][0], dp[j][1] + 1); + } + } + } + return max(dp[numsSize - 1][0], dp[numsSize - 1][1]); +} +``` + + ### TypeScript From d4f89d1b0eec92115d0270630c77a09803b45703 Mon Sep 17 00:00:00 2001 From: programmercarl <826123027@qq.com> Date: Wed, 20 Jul 2022 09:45:11 +0800 Subject: [PATCH 128/136] =?UTF-8?q?=E4=BC=98=E5=8C=96=E5=8E=9F=E9=A2=98?= =?UTF-8?q?=E8=A7=A3=EF=BC=8C=E6=B7=BB=E5=8A=A0=E9=83=A8=E5=88=86=E9=A2=98?= =?UTF-8?q?=E7=9B=AE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0001.两数之和.md | 86 +++-- problems/0015.三数之和.md | 96 +++++- problems/0018.四数之和.md | 37 +-- problems/0024.两两交换链表中的节点.md | 6 +- ...排序数组中查找元素的第一个和最后一个位置.md | 2 + problems/0056.合并区间.md | 3 - problems/0122.买卖股票的最佳时机II.md | 4 +- problems/0142.环形链表II.md | 2 + problems/0202.快乐数.md | 47 --- problems/0206.翻转链表.md | 2 + problems/0209.长度最小的子数组.md | 2 +- problems/0216.组合总和III.md | 2 +- problems/0242.有效的字母异位词.md | 2 + problems/0349.两个数组的交集.md | 37 ++- problems/0376.摆动序列.md | 12 +- problems/0392.判断子序列.md | 6 +- problems/0454.四数相加II.md | 13 +- problems/0647.回文子串.md | 10 +- problems/0704.二分查找.md | 2 +- problems/0707.设计链表.md | 2 + problems/0718.最长重复子数组.md | 2 +- problems/0797.所有可能的路径.md | 296 ++++++++++++++++++ problems/0841.钥匙和房间.md | 229 ++++++++++---- problems/1791.找出星型图的中心节点.md | 73 +++++ problems/1971.寻找图中是否存在路径.md | 123 ++++++++ problems/动态规划总结篇.md | 2 +- problems/面试题02.07.链表相交.md | 4 +- 27 files changed, 896 insertions(+), 206 deletions(-) create mode 100644 problems/0797.所有可能的路径.md create mode 100644 problems/1791.找出星型图的中心节点.md create mode 100644 problems/1971.寻找图中是否存在路径.md diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index fb3c1d45..055f8940 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -24,7 +24,9 @@ ## 思路 -很明显暴力的解法是两层for循环查找，时间复杂度是$O(n^2)$。 +建议看一下我录的这期视频：[梦开始的地方，Leetcode：1.两数之和](https://www.bilibili.com/video/BV1aT41177mK)，结合本题解来学习，事半功倍。 + +很明显暴力的解法是两层for循环查找，时间复杂度是O(n^2)。 建议大家做这道题目之前，先做一下这两道 * [242. 有效的字母异位词](https://www.programmercarl.com/0242.有效的字母异位词.html) @@ -32,7 +34,16 @@ [242. 有效的字母异位词](https://www.programmercarl.com/0242.有效的字母异位词.html) 这道题目是用数组作为哈希表来解决哈希问题，[349. 两个数组的交集](https://www.programmercarl.com/0349.两个数组的交集.html)这道题目是通过set作为哈希表来解决哈希问题。 -本题呢，则要使用map，那么来看一下使用数组和set来做哈希法的局限。 + +首先我在强调一下 **什么时候使用哈希法**，当我们需要查询一个元素是否出现过，或者一个元素是否在集合里的时候，就要第一时间想到哈希法。 + +本题呢，我就需要一个集合来存放我们遍历过的元素，然后在遍历数组的时候去询问这个集合，某元素是否遍历过，也就是 是否出现在这个集合。 + +那么我们就应该想到使用哈希法了。 + +因为本地，我们不仅要知道元素有没有遍历过，还有知道这个元素对应的下标，**需要使用 key value结构来存放，key来存元素，value来存下标，那么使用map正合适**。 + +再来看一下使用数组和set来做哈希法的局限。 * 数组的大小是受限制的，而且如果元素很少，而哈希值太大会造成内存空间的浪费。 * set是一个集合，里面放的元素只能是一个key，而两数之和这道题目，不仅要判断y是否存在而且还要记录y的下标位置，因为要返回x 和 y的下标。所以set 也不能用。 @@ -43,20 +54,38 @@ C++中map，有三种类型： |映射 |底层实现 | 是否有序 |数值是否可以重复 | 能否更改数值|查询效率 |增删效率| |---|---| --- |---| --- | --- | ---| -|std::map |红黑树 |key有序 |key不可重复 |key不可修改 | $O(\log n)$|$O(\log n)$ | -|std::multimap | 红黑树|key有序 | key可重复 | key不可修改|$O(\log n)$ |$O(\log n)$ | -|std::unordered_map |哈希表 | key无序 |key不可重复 |key不可修改 |$O(1)$ | $O(1)$| +|std::map |红黑树 |key有序 |key不可重复 |key不可修改 | O(log n)|O(log n) | +|std::multimap | 红黑树|key有序 | key可重复 | key不可修改|O(log n) |O(log n) | +|std::unordered_map |哈希表 | key无序 |key不可重复 |key不可修改 |O(1) | O(1)| std::unordered_map 底层实现为哈希表，std::map 和std::multimap 的底层实现是红黑树。 同理，std::map 和std::multimap 的key也是有序的（这个问题也经常作为面试题，考察对语言容器底层的理解）。 更多哈希表的理论知识请看[关于哈希表，你该了解这些！](https://www.programmercarl.com/哈希表理论基础.html)。 -**这道题目中并不需要key有序，选择std::unordered_map 效率更高！** +**这道题目中并不需要key有序，选择std::unordered_map 效率更高！** 使用其他语言的录友注意了解一下自己所用语言的数据结构就行。 -解题思路动画如下： +接下来需要明确两点： -![](https://code-thinking.cdn.bcebos.com/gifs/1.两数之和.gif) +* **map用来做什么** +* **map中key和value分别表示什么** +map目的用来存放我们访问过的元素，因为遍历数组的时候，需要记录我们之前遍历过哪些元素和对应的下表，这样才能找到与当前元素相匹配的（也就是相加等于target） + +接下来是map中key和value分别表示什么。 + +这道题 我们需要 给出一个元素，判断这个元素是否出现过，如果出现过，返回这个元素的下标。 + +那么判断元素是否出现，这个元素就要作为key，所以数组中的元素作为key，有key对应的就是value，value用来存下标。 + +所以 map中的存储结构为 {key：数据元素，value：数组元素对应的下表}。 + +在遍历数组的时候，只需要向map去查询是否有和目前遍历元素比配的数值，如果有，就找到的匹配对，如果没有，就把目前遍历的元素放进map中，因为map存放的就是我们访问过的元素。 + +过程如下： + +![过程一](https://code-thinking-1253855093.file.myqcloud.com/pics/20220711202638.png) + +![过程二](https://code-thinking-1253855093.file.myqcloud.com/pics/20220711202708.png) C++代码： @@ -66,18 +95,31 @@ public: vector twoSum(vector& nums, int target) { std::unordered_map map; for(int i = 0; i < nums.size(); i++) { - auto iter = map.find(target - nums[i]); + // 遍历当前元素，并在map中寻找是否有匹配的key + auto iter = map.find(target - nums[i]); if(iter != map.end()) { return {iter->second, i}; } - map.insert(pair(nums[i], i)); + // 如果没找到匹配对，就把访问过的元素和下标加入到map中 + map.insert(pair(nums[i], i)); } return {}; } }; ``` +## 总结 +本题其实有四个重点： + +* 为什么会想到用哈希表 +* 哈希表为什么用map +* 本题map是用来存什么的 +* map中的key和value用来存什么的 + +把这四点想清楚了，本题才算是理解透彻了。 + +很多录友把这道题目 通过了，但都没想清楚map是用来做什么的，以至于对代码的理解其实是 一知半解的。 ## 其他语言版本 @@ -250,30 +292,6 @@ func twoSum(_ nums: [Int], _ target: Int) -> [Int] { } ``` -PHP: -```php -class Solution { - /** - * @param Integer[] $nums - * @param Integer $target - * @return Integer[] - */ - function twoSum($nums, $target) { - if (count($nums) == 0) { - return []; - } - $table = []; - for ($i = 0; $i < count($nums); $i++) { - $temp = $target - $nums[$i]; - if (isset($table[$temp])) { - return [$table[$temp], $i]; - } - $table[$nums[$i]] = $i; - } - return []; - } -} -``` Scala: ```scala diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md index e6dc82dd..5a4168d1 100644 --- a/problems/0015.三数之和.md +++ b/problems/0015.三数之和.md @@ -39,7 +39,7 @@ 去重的过程不好处理，有很多小细节，如果在面试中很难想到位。 -时间复杂度可以做到$O(n^2)$，但还是比较费时的，因为不好做剪枝操作。 +时间复杂度可以做到O(n^2)，但还是比较费时的，因为不好做剪枝操作。 大家可以尝试使用哈希法写一写，就知道其困难的程度了。 @@ -85,7 +85,7 @@ public: **其实这道题目使用哈希法并不十分合适**，因为在去重的操作中有很多细节需要注意，在面试中很难直接写出没有bug的代码。 -而且使用哈希法 在使用两层for循环的时候，能做的剪枝操作很有限，虽然时间复杂度是$O(n^2)$，也是可以在leetcode上通过，但是程序的执行时间依然比较长 。 +而且使用哈希法 在使用两层for循环的时候，能做的剪枝操作很有限，虽然时间复杂度是O(n^2)，也是可以在leetcode上通过，但是程序的执行时间依然比较长 。 接下来我来介绍另一个解法：双指针法，**这道题目使用双指针法 要比哈希法高效一些**，那么来讲解一下具体实现的思路。 @@ -101,7 +101,7 @@ public: 如果 nums[i] + nums[left] + nums[right] < 0 说明 此时 三数之和小了，left 就向右移动，才能让三数之和大一些，直到left与right相遇为止。 -时间复杂度：$O(n^2)$。 +时间复杂度：O(n^2)。 C++代码代码如下： @@ -118,13 +118,13 @@ public: if (nums[i] > 0) { return result; } - // 错误去重方法，将会漏掉-1,-1,2 这种情况 + // 错误去重a方法，将会漏掉-1,-1,2 这种情况 /* if (nums[i] == nums[i + 1]) { continue; } */ - // 正确去重方法 + // 正确去重a方法 if (i > 0 && nums[i] == nums[i - 1]) { continue; } @@ -136,17 +136,11 @@ public: while (right > left && nums[right] == nums[right - 1]) right--; while (right > left && nums[left] == nums[left + 1]) left++; */ - if (nums[i] + nums[left] + nums[right] > 0) { - right--; - // 当前元素不合适了，可以去重 - while (left < right && nums[right] == nums[right + 1]) right--; - } else if (nums[i] + nums[left] + nums[right] < 0) { - left++; - // 不合适，去重 - while (left < right && nums[left] == nums[left - 1]) left++; - } else { + if (nums[i] + nums[left] + nums[right] > 0) right--; + else if (nums[i] + nums[left] + nums[right] < 0) left++; + else { result.push_back(vector{nums[i], nums[left], nums[right]}); - // 去重逻辑应该放在找到一个三元组之后 + // 去重逻辑应该放在找到一个三元组之后，对b 和 c去重 while (right > left && nums[right] == nums[right - 1]) right--; while (right > left && nums[left] == nums[left + 1]) left++; @@ -162,6 +156,78 @@ public: }; ``` +## 去重逻辑的思考 + +### a的去重 + +说道去重，其实主要考虑三个数的去重。 a, b ,c, 对应的就是 nums[i]，nums[left]，nums[right] + +a 如果重复了怎么办，a是nums里遍历的元素，那么应该直接跳过去。 + +但这里有一个问题，是判断 nums[i] 与 nums[i + 1]是否相同，还是判断 nums[i] 与 nums[i-1] 是否相同。 + +有同学可能想，这不都一样吗。 + +其实不一样！ + +都是和 nums[i]进行比较，是比较它的前一个，还是比较他的后一个。 + +如果我们的写法是 这样： + +```C++ +if (nums[i] == nums[i + 1]) { // 去重操作 + continue; +} +``` + +那就我们就把 三元组中出现重复元素的情况直接pass掉了。 例如{-1, -1 ,2} 这组数据，当遍历到第一个-1 的时候，判断 下一个也是-1，那这组数据就pass了。 + +**我们要做的是 不能有重复的三元组，但三元组内的元素是可以重复的！** + +所以这里是有两个重复的维度。 + +那么应该这么写： + +```C++ +if (i > 0 && nums[i] == nums[i - 1]) { + continue; +} +``` + +这么写就是当前使用 nums[i]，我们判断前一位是不是一样的元素，在看 {-1, -1 ,2} 这组数据，当遍历到 第一个 -1 的时候，只要前一位没有-1，那么 {-1, -1 ,2} 这组数据一样可以收录到 结果集里。 + +这是一个非常细节的思考过程。 + +### b与c的去重 + +很多同学写本题的时候，去重的逻辑多加了 对right 和left 的去重：（代码中注释部分） + +```C++ +while (right > left) { + if (nums[i] + nums[left] + nums[right] > 0) { + right--; + // 去重 right + while (left < right && nums[right] == nums[right + 1]) right--; + } else if (nums[i] + nums[left] + nums[right] < 0) { + left++; + // 去重 left + while (left < right && nums[left] == nums[left - 1]) left++; + } else { + } +} +``` + +但细想一下，这种去重其实对提升程序运行效率是没有帮助的。 + +拿right去重为例，即使不加这个去重逻辑，依然根据 `while (right > left) ` 和 `if (nums[i] + nums[left] + nums[right] > 0)` 去完成right-- 的操作。 + +多加了 ` while (left < right && nums[right] == nums[right + 1]) right--;` 这一行代码，其实就是把 需要执行的逻辑提前执行了，但并没有减少 判断的逻辑。 + +最直白的思考过程，就是right还是一个数一个数的减下去的，所以在哪里减的都是一样的。 + +所以这种去重 是可以不加的。 仅仅是 把去重的逻辑提前了而已。 + + # 思考题 diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 2146a114..7989ad8f 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -35,11 +35,11 @@ [15.三数之和](https://programmercarl.com/0015.三数之和.html)的双指针解法是一层for循环num[i]为确定值，然后循环内有left和right下标作为双指针，找到nums[i] + nums[left] + nums[right] == 0。 -四数之和的双指针解法是两层for循环nums[k] + nums[i]为确定值，依然是循环内有left和right下标作为双指针，找出nums[k] + nums[i] + nums[left] + nums[right] == target的情况，三数之和的时间复杂度是$O(n^2)$，四数之和的时间复杂度是$O(n^3)$ 。 +四数之和的双指针解法是两层for循环nums[k] + nums[i]为确定值，依然是循环内有left和right下标作为双指针，找出nums[k] + nums[i] + nums[left] + nums[right] == target的情况，三数之和的时间复杂度是O(n^2)，四数之和的时间复杂度是O(n^3) 。 那么一样的道理，五数之和、六数之和等等都采用这种解法。 -对于[15.三数之和](https://programmercarl.com/0015.三数之和.html)双指针法就是将原本暴力$O(n^3)$的解法，降为$O(n^2)$的解法，四数之和的双指针解法就是将原本暴力$O(n^4)$的解法，降为$O(n^3)$的解法。 +对于[15.三数之和](https://programmercarl.com/0015.三数之和.html)双指针法就是将原本暴力O(n^3)的解法，降为O(n^2)的解法，四数之和的双指针解法就是将原本暴力O(n^4)的解法，降为O(n^3)的解法。 之前我们讲过哈希表的经典题目：[454.四数相加II](https://programmercarl.com/0454.四数相加II.html)，相对于本题简单很多，因为本题是要求在一个集合中找出四个数相加等于target，同时四元组不能重复。 @@ -47,14 +47,13 @@ 我们来回顾一下，几道题目使用了双指针法。 -双指针法将时间复杂度：$O(n^2)$的解法优化为 $O(n)$的解法。也就是降一个数量级，题目如下： +双指针法将时间复杂度：O(n^2)的解法优化为 O(n)的解法。也就是降一个数量级，题目如下： * [27.移除元素](https://programmercarl.com/0027.移除元素.html) * [15.三数之和](https://programmercarl.com/0015.三数之和.html) * [18.四数之和](https://programmercarl.com/0018.四数之和.html) - -操作链表： +链表相关双指针题目： * [206.反转链表](https://programmercarl.com/0206.翻转链表.html) * [19.删除链表的倒数第N个节点](https://programmercarl.com/0019.删除链表的倒数第N个节点.html) @@ -72,21 +71,21 @@ public: vector> result; sort(nums.begin(), nums.end()); for (int k = 0; k < nums.size(); k++) { - // 剪枝处理 - if (nums[k] > target && (nums[k] >= 0 || target >= 0)) { + // 剪枝处理 + if (nums[k] > target && nums[k] >= 0 && target >= 0) { break; // 这里使用break，统一通过最后的return返回 } - // 去重 + // 对nums[k]去重 if (k > 0 && nums[k] == nums[k - 1]) { continue; } for (int i = k + 1; i < nums.size(); i++) { - // 2级剪枝处理 - if (nums[k] + nums[i] > target && (nums[k] + nums[i] >= 0 || target >= 0)) { - break; - } - - // 正确去重方法 + // 2级剪枝处理 + if (nums[k] + nums[i] > target && nums[k] + nums[i] >= 0 && target >= 0) { + break; + } + + // 对nums[i]去重 if (i > k + 1 && nums[i] == nums[i - 1]) { continue; } @@ -94,18 +93,14 @@ public: int right = nums.size() - 1; while (right > left) { // nums[k] + nums[i] + nums[left] + nums[right] > target 会溢出 - if (nums[k] + nums[i] > target - (nums[left] + nums[right])) { + if ((long) nums[k] + nums[i] + nums[left] + nums[right] > target) { right--; - // 当前元素不合适了，可以去重 - while (left < right && nums[right] == nums[right + 1]) right--; // nums[k] + nums[i] + nums[left] + nums[right] < target 会溢出 - } else if (nums[k] + nums[i] < target - (nums[left] + nums[right])) { + } else if ((long) nums[k] + nums[i] + nums[left] + nums[right] < target) { left++; - // 不合适，去重 - while (left < right && nums[left] == nums[left - 1]) left++; } else { result.push_back(vector{nums[k], nums[i], nums[left], nums[right]}); - // 去重逻辑应该放在找到一个四元组之后 + // 对nums[left]和nums[right]去重 while (right > left && nums[right] == nums[right - 1]) right--; while (right > left && nums[left] == nums[left + 1]) left++; diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md index e636bfff..550a886e 100644 --- a/problems/0024.两两交换链表中的节点.md +++ b/problems/0024.两两交换链表中的节点.md @@ -18,6 +18,8 @@ ## 思路 +针对本题重点难点，我录制了B站讲解视频，[帮你把链表细节学清楚！ | LeetCode：24. 两两交换链表中的节点](https://www.bilibili.com/video/BV1YT411g7br)，相信结合视频在看本篇题解，更有助于大家对链表的理解。 + 这道题目正常模拟就可以了。 建议使用虚拟头结点，这样会方便很多，要不然每次针对头结点（没有前一个指针指向头结点），还要单独处理。 @@ -63,8 +65,8 @@ public: }; ``` -* 时间复杂度：$O(n)$ -* 空间复杂度：$O(1)$ +* 时间复杂度：O(n) +* 空间复杂度：O(1) ## 拓展 diff --git a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md index 260462c2..bf9493c9 100644 --- a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md +++ b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md @@ -7,6 +7,8 @@ # 34. 在排序数组中查找元素的第一个和最后一个位置 +[题目链接](https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/) + 给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。 如果数组中不存在目标值 target，返回 [-1, -1]。 diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index 34d8dd82..757896c1 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -22,9 +22,6 @@ * 解释: 区间 [1,4] 和 [4,5] 可被视为重叠区间。 * 注意：输入类型已于2019年4月15日更改。 请重置默认代码定义以获取新方法签名。 -提示： - -* intervals[i][0] <= intervals[i][1] ## 思路 diff --git a/problems/0122.买卖股票的最佳时机II.md b/problems/0122.买卖股票的最佳时机II.md index b9fa8386..ae9744d1 100644 --- a/problems/0122.买卖股票的最佳时机II.md +++ b/problems/0122.买卖股票的最佳时机II.md @@ -91,8 +91,8 @@ public: }; ``` -* 时间复杂度：$O(n)$ -* 空间复杂度：$O(1)$ +* 时间复杂度：O(n) +* 空间复杂度：O(1) ### 动态规划 diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md index 6b7c7e66..254f9db0 100644 --- a/problems/0142.环形链表II.md +++ b/problems/0142.环形链表II.md @@ -24,6 +24,8 @@ ## 思路 +为了易于大家理解，我录制讲解视频：[B站：把环形链表讲清楚！ ](https://www.bilibili.com/video/BV1if4y1d7ob)。结合视频在看本篇题解，事半功倍。 + 这道题目，不仅考察对链表的操作，而且还需要一些数学运算。 主要考察两知识点： diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 0bea0c72..0054582c 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -338,53 +338,6 @@ static inline int calcSquareSum(int num) { return sum; } -#define HASH_TABLE_SIZE (32) - -bool isHappy(int n){ - int sum = n; - int index = 0; - bool bHappy = false; - bool bExit = false; - /* allocate the memory for hash table with chaining method*/ - HashNode ** hashTable = (HashNode **)calloc(HASH_TABLE_SIZE, sizeof(HashNode)); - - while(bExit == false) { - /* check if n has been calculated */ - index = hash(n, HASH_TABLE_SIZE); - - HashNode ** p = hashTable + index; - - while((*p) && (bExit == false)) { - /* Check if this num was calculated, if yes, this will be endless loop */ - if((*p)->key == n) { - bHappy = false; - bExit = true; - } - /* move to next node of the same index */ - p = &((*p)->next); - } - - /* put n intot hash table */ - HashNode * newNode = (HashNode *)malloc(sizeof(HashNode)); - newNode->key = n; - newNode->next = NULL; - - *p = newNode; - - sum = calcSquareSum(n); - if(sum == 1) { - bHappy = true; - bExit = true; - } - else { - n = sum; - - } - } - - return bHappy; -} -``` Scala: ```scala diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index 24ec7b94..c412e840 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -19,6 +19,8 @@ # 思路 +本题我录制了B站视频，[帮你拿下反转链表 | LeetCode：206.反转链表](https://www.bilibili.com/video/BV1nB4y1i7eL)，相信结合视频在看本篇题解，更有助于大家对链表的理解。 + 如果再定义一个新的链表，实现链表元素的反转，其实这是对内存空间的浪费。 其实只需要改变链表的next指针的指向，直接将链表反转 ，而不用重新定义一个新的链表，如图所示: diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index e2eb378e..b374aaec 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -19,7 +19,7 @@ # 思路 -为了易于大家理解，我特意录制了[拿下滑动窗口！ | LeetCode 209 长度最小的子数组](https://www.bilibili.com/video/BV1tZ4y1q7XE) +为了易于大家理解，我特意录制了B站视频[拿下滑动窗口！ | LeetCode 209 长度最小的子数组](https://www.bilibili.com/video/BV1tZ4y1q7XE) ## 暴力解法 diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md index 0ace0fd5..a8a6567d 100644 --- a/problems/0216.组合总和III.md +++ b/problems/0216.组合总和III.md @@ -212,7 +212,7 @@ public: # 总结 -开篇就介绍了本题与[回溯算法：求组合问题！](https://programmercarl.com/0077.组合.html)的区别，相对来说加了元素总和的限制，如果做完[回溯算法：求组合问题！](https://programmercarl.com/0077.组合.html)再做本题在合适不过。 +开篇就介绍了本题与[77.组合](https://programmercarl.com/0077.组合.html)的区别，相对来说加了元素总和的限制，如果做完[77.组合](https://programmercarl.com/0077.组合.html)再做本题在合适不过。 分析完区别，依然把问题抽象为树形结构，按照回溯三部曲进行讲解，最后给出剪枝的优化。 diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index 8f4b5ae2..0437330b 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -27,6 +27,8 @@ ## 思路 +本题B站视频讲解版：[学透哈希表，数组使用有技巧！Leetcode：242.有效的字母异位词](https://www.bilibili.com/video/BV1YG411p7BA) + 先看暴力的解法，两层for循环，同时还要记录字符是否重复出现，很明显时间复杂度是 O(n^2)。 暴力的方法这里就不做介绍了，直接看一下有没有更优的方式。 diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 98518647..c1d3515b 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -24,6 +24,8 @@ ## 思路 +关于本题，我录制了讲解视频：[学透哈希表，set使用有技巧！Leetcode：349. 两个数组的交集](https://www.bilibili.com/video/BV1ba411S7wu)，看视频配合题解，事半功倍。 + 这道题目，主要要学会使用一种哈希数据结构：unordered_set，这个数据结构可以解决很多类似的问题。 注意题目特意说明：**输出结果中的每个元素一定是唯一的，也就是说输出的结果的去重的， 同时可以不考虑输出结果的顺序** @@ -48,7 +50,8 @@ std::set和std::multiset底层实现都是红黑树，std::unordered_set的底 思路如图所示： -![set哈希法](https://img-blog.csdnimg.cn/2020080918570417.png) + +![set哈希法](https://code-thinking-1253855093.file.myqcloud.com/pics/20220707173513.png) C++代码如下： @@ -56,7 +59,7 @@ C++代码如下： class Solution { public: vector intersection(vector& nums1, vector& nums2) { - unordered_set result_set; // 存放结果 + unordered_set result_set; // 存放结果，之所以用set是为了给结果集去重 unordered_set nums_set(nums1.begin(), nums1.end()); for (int num : nums2) { // 发现nums2的元素 在nums_set里又出现过 @@ -77,6 +80,36 @@ public: 不要小瞧 这个耗时，在数据量大的情况，差距是很明显的。 +## 后记 + +本题后面 力扣改了 题目描述 和 后台测试数据，增添了 数值范围： + +* 1 <= nums1.length, nums2.length <= 1000 +* 0 <= nums1[i], nums2[i] <= 1000 + +所以就可以 使用数组来做哈希表了， 因为数组都是 1000以内的。 + +对应C++代码如下： + +```c++ +class Solution { +public: + vector intersection(vector& nums1, vector& nums2) { + unordered_set result_set; // 存放结果，之所以用set是为了给结果集去重 + int hash[1005] = {0}; // 默认数值为0 + for (int num : nums1) { // nums1中出现的字母在hash数组中做记录 + hash[num] = 1; + } + for (int num : nums2) { // nums2中出现话，result记录 + if (hash[num] == 1) { + result_set.insert(num); + } + } + return vector(result_set.begin(), result_set.end()); + } +}; +``` + ## 其他语言版本 diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index 00f8f70c..e8db980d 100644 --- a/problems/0376.摆动序列.md +++ b/problems/0376.摆动序列.md @@ -88,8 +88,8 @@ public: }; ``` -* 时间复杂度：$O(n)$ -* 空间复杂度：$O(1)$ +* 时间复杂度：O(n) +* 空间复杂度：O(1) ## 思路2（动态规划） @@ -138,8 +138,8 @@ public: }; ``` -* 时间复杂度：$O(n^2)$ -* 空间复杂度：$O(n)$ +* 时间复杂度：O(n^2) +* 空间复杂度：O(n) **进阶** @@ -149,9 +149,9 @@ public: * 每次更新`dp[i][1]`，则在`tree2`的`nums[i]`位置值更新为`dp[i][1]` * 则dp转移方程中就没有必要j从0遍历到i-1，可以直接在线段树中查询指定区间的值即可。 -时间复杂度：$O(n\log n)$ +时间复杂度：O(nlog n) -空间复杂度：$O(n)$ +空间复杂度：O(n) ## 总结 diff --git a/problems/0392.判断子序列.md b/problems/0392.判断子序列.md index 9a26d639..dedeb0ad 100644 --- a/problems/0392.判断子序列.md +++ b/problems/0392.判断子序列.md @@ -31,7 +31,7 @@ ## 思路 -（这道题可以用双指针的思路来实现，时间复杂度就是$O(n)$） +（这道题可以用双指针的思路来实现，时间复杂度就是O(n)） 这道题应该算是编辑距离的入门题目，因为从题意中我们也可以发现，只需要计算删除的情况，不用考虑增加和替换的情况。 @@ -122,8 +122,8 @@ public: }; ``` -* 时间复杂度：$O(n × m)$ -* 空间复杂度：$O(n × m)$ +* 时间复杂度：O(n × m) +* 空间复杂度：O(n × m) ## 总结 diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index d22e2335..b735c106 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -18,14 +18,19 @@ **例如:** 输入: -A = [ 1, 2] -B = [-2,-1] -C = [-1, 2] -D = [ 0, 2] +* A = [ 1, 2] +* B = [-2,-1] +* C = [-1, 2] +* D = [ 0, 2] + 输出: + 2 + **解释:** + 两个元组如下: + 1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0 2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0 diff --git a/problems/0647.回文子串.md b/problems/0647.回文子串.md index c0b34e8a..fa9bf7b8 100644 --- a/problems/0647.回文子串.md +++ b/problems/0647.回文子串.md @@ -32,7 +32,7 @@ 两层for循环，遍历区间起始位置和终止位置，然后判断这个区间是不是回文。 -时间复杂度：$O(n^3)$ +时间复杂度：O(n^3) ## 动态规划 @@ -171,8 +171,8 @@ public: }; ``` -* 时间复杂度：$O(n^2)$ -* 空间复杂度：$O(n^2)$ +* 时间复杂度：O(n^2) +* 空间复杂度：O(n^2) ## 双指针法 @@ -213,8 +213,8 @@ public: }; ``` -* 时间复杂度：$O(n^2)$ -* 空间复杂度：$O(1)$ +* 时间复杂度：O(n^2) +* 空间复杂度：O(1) ## 其他语言版本 diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index 6a37e4d1..432305e2 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -36,7 +36,7 @@ ## 思路 -为了易于大家理解，我还录制了视频，可以看这里：[手把手带你撕出正确的二分法](https://www.bilibili.com/video/BV1fA4y1o715) +为了易于大家理解，我还录制了视频，可以看这里：[B站：手把手带你撕出正确的二分法](https://www.bilibili.com/video/BV1fA4y1o715) **这道题目的前提是数组为有序数组**，同时题目还强调**数组中无重复元素**，因为一旦有重复元素，使用二分查找法返回的元素下标可能不是唯一的，这些都是使用二分法的前提条件，当大家看到题目描述满足如上条件的时候，可要想一想是不是可以用二分法了。 diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 6ee11eef..b0385ce1 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -26,6 +26,8 @@ # 思路 +为了方便大家理解，我特意录制了视频：[帮你把链表操作学个通透！LeetCode：707.设计链表](https://www.bilibili.com/video/BV1FU4y1X7WD)，结合视频在看本题解，事半功倍。 + 如果对链表的基础知识还不太懂，可以看这篇文章：[关于链表，你该了解这些！](https://programmercarl.com/链表理论基础.html) 如果对链表的虚拟头结点不清楚，可以看这篇文章：[链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html) diff --git a/problems/0718.最长重复子数组.md b/problems/0718.最长重复子数组.md index 18007b70..6de8b80f 100644 --- a/problems/0718.最长重复子数组.md +++ b/problems/0718.最长重复子数组.md @@ -31,7 +31,7 @@ B: [3,2,1,4,7] 1. 确定dp数组（dp table）以及下标的含义 -dp[i][j] ：以下标i - 1为结尾的A，和以下标j - 1为结尾的B，最长重复子数组长度为dp[i][j]。 +dp[i][j] ：以下标i - 1为结尾的A，和以下标j - 1为结尾的B，最长重复子数组长度为dp[i][j]。 （**特别注意**： “以下标i - 1为结尾的A” 标明一定是 以A[i-1]为结尾的字符串 ） 此时细心的同学应该发现，那dp[0][0]是什么含义呢？总不能是以下标-1为结尾的A数组吧。 diff --git a/problems/0797.所有可能的路径.md b/problems/0797.所有可能的路径.md new file mode 100644 index 00000000..b990053e --- /dev/null +++ b/problems/0797.所有可能的路径.md @@ -0,0 +1,296 @@ + +看一下 算法4，深搜是怎么讲的 + +# 797.所有可能的路径 + +本题是一道 原汁原味的 深度优先搜索（dfs）模板题，那么用这道题目 来讲解 深搜最合适不过了。 + +接下来给大家详细讲解dfs： + +## dfs 与 bfs 区别 + +先来了解dfs的过程，很多录友可能对dfs（深度优先搜索），bfs（广度优先搜索）分不清。 + +先给大家说一下两者大概的区别： + +* dfs是可一个方向去搜，不到黄河不回头，直到遇到绝境了，搜不下去了，在换方向（换方向的过程就涉及到了回溯）。 +* bfs是先把本节点所连接的所有节点遍历一遍，走到下一个节点的时候，再把连接节点的所有节点遍历一遍，搜索方向更像是广度，四面八方的搜索过程。 + +当然以上讲的是，大体可以这么理解，接下来 我们详细讲解dfs，（bfs在用单独一篇文章详细讲解） + +## dfs 搜索过程 + +上面说道dfs是可一个方向搜，不到黄河不回头。 那么我们来举一个例子。 + +如图一，是一个无向图，我们要搜索从节点1到节点6的所有路径。 + +![图一](https://code-thinking-1253855093.file.myqcloud.com/pics/20220707093643.png) + +那么dfs搜索的第一条路径是这样的： （假设第一次延默认方向，就找到了节点6），图二 + +![图二](https://code-thinking-1253855093.file.myqcloud.com/pics/20220707093807.png) + +此时我们找到了节点6，（遇到黄河了，是不是应该回头了），那么应该再去搜索其他方向了。 如图三： + +![图三](https://code-thinking-1253855093.file.myqcloud.com/pics/20220707094011.png) + +路径2撤销了，改变了方向，走路径3（红色线）， 接着也找到终点6。 那么撤销路径2，改为路径3，在dfs中其实就是回溯的过程（这一点很重要，很多录友，都不理解dfs代码中回溯是用来干什么的） + +又找到了一条从节点1到节点6的路径，又到黄河了，此时再回头，下图图四中，路径4撤销（回溯的过程），改为路径5。 + +![图四](https://code-thinking-1253855093.file.myqcloud.com/pics/20220707094322.png) + +又找到了一条从节点1到节点6的路径，又到黄河了，此时再回头，下图图五，路径6撤销（回溯的过程），改为路径7，路径8 和 路径7，路径9， 结果发现死路一条，都走到了自己走过的节点。 + +![图五](https://code-thinking-1253855093.file.myqcloud.com/pics/20220707094813.png) + +那么节点2所连接路径和节点3所链接的路径 都走过了，撤销路径只能向上回退，去选择撤销当初节点4的选择，也就是撤销路径5，改为路径10 。 如图图六： + +![图六](https://code-thinking-1253855093.file.myqcloud.com/pics/20220707095232.png) + + +上图演示中，其实我并没有把 所有的 从节点1 到节点6的dfs（深度优先搜索）的过程都画出来，那样太冗余了，但 已经把dfs 关键的地方都涉及到了，关键就两点： + +* 搜索方向，是认准一个方向搜，直到碰壁之后在换方向 +* 换方向是撤销原路径，改为节点链接的下一个路径，回溯的过程。 + + +## 代码框架 + +正式因为dfs搜索可一个方向，并需要回溯，所以用递归的方式来实现是最方便的。 + +很多录友对回溯很陌生，建议先看看码随想录，[回溯算法章节](https://programmercarl.com/回溯算法理论基础.html)。 + +有递归的地方就有回溯，那么回溯在哪里呢？ + +就地递归函数的下面，例如如下代码： +``` +void dfs(参数) { + 处理节点 + dfs(图，选择的节点); // 递归 + 回溯，撤销处理结果 +} +``` + +可以看到回溯操作就在递归函数的下面，递归和回溯是相辅相成的。 + +在讲解[二叉树章节](https://programmercarl.com/二叉树理论基础.html)的时候，二叉树的递归法其实就是dfs，而二叉树的迭代法，就是bfs（广度优先搜索） + +所以**dfs，bfs其实是基础搜索算法，也广泛应用与其他数据结构与算法中**。 + +我们在回顾一下[回溯法](https://programmercarl.com/回溯算法理论基础.html)的代码框架： + +``` +void backtracking(参数) { + if (终止条件) { + 存放结果; + return; + } + for (选择：本层集合中元素（树中节点孩子的数量就是集合的大小）) { + 处理节点; + backtracking(路径，选择列表); // 递归 + 回溯，撤销处理结果 + } +} + +``` + +回溯算法，其实就是dfs的过程，这里给出dfs的代码框架： + +``` +void dfs(参数) { + if (终止条件) { + 存放结果; + return; + } + + for (选择：本节点所连接的其他节点) { + 处理节点; + dfs(图，选择的节点); // 递归 + 回溯，撤销处理结果 + } +} + +``` + +可以发现dfs的代码框架和回溯算法的代码框架是差不多的。 + +下面我在用 深搜三部曲，来解读 dfs的代码框架。 + +## 深搜三部曲 + +在 [二叉树递归讲解](https://programmercarl.com/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E9%80%92%E5%BD%92%E9%81%8D%E5%8E%86.html)中，给出了递归三部曲。 + +[回溯算法](https://programmercarl.com/回溯算法理论基础.html)讲解中，给出了 回溯三部曲。 + +其实深搜也是一样的，深搜三部曲如下： + +1. 确认递归函数，参数 + +``` +void dfs(参数) +``` + +通常我们递归的时候，我们递归搜索需要了解哪些参数，其实也可以在写递归函数的时候，发现需要什么参数，再去补充就可以。 + +一般情况，深搜需要 二维数组数组结构保存所有路径，需要一维数组保存单一路径，这种保存结果的数组，我们可以定义一个全局遍历，避免让我们的函数参数过多。 + +例如这样： + +``` +vector> result; // 保存符合条件的所有路径 +vector path; // 起点到终点的路径 +void dfs (图，目前搜索的节点) +``` + +但这种写法看个人习惯，不强求。 + +2. 确认终止条件 + +终止条件很重要，很多同学写dfs的时候，之所以容易死循环，栈溢出等等这些问题，都是因为终止条件没有想清楚。 + +``` +if (终止条件) { + 存放结果; + return; +} +``` + +终止添加不仅是结束本层递归，同时也是我们收获结果的时候。 + +3. 处理目前搜索节点出发的路径 + +一般这里就是一个for循环的操作，去遍历 目前搜索节点 所能到的所有节点。 + +``` +for (选择：本节点所连接的其他节点) { + 处理节点; + dfs(图，选择的节点); // 递归 + 回溯，撤销处理结果 +} +``` + +不少录友疑惑的地方，都是 dfs代码框架中for循环里分明已经处理节点了，那么 dfs函数下面 为什么还要撤销的呢。 + +如图七所示， 路径2 已经走到了 目的地节点6，那么 路径2 是如何撤销，然后改为 路径3呢？ 其实这就是 回溯的过程，撤销路径2，走换下一个方向。 + +![图七](https://code-thinking-1253855093.file.myqcloud.com/pics/20220708093544.png) + + +## 总结 + +我们讲解了，dfs 和 bfs的大体区别（bfs详细过程下篇来讲），dfs的搜索过程以及代码框架。 + +最后还有 深搜三部曲来解读这份代码框架。 + +以上如果大家都能理解了，其实搜索的代码就很好写，具体题目套用具体场景就可以了。 + +## 797. 所有可能的路径 + +### 思路 + +1. 确认递归函数，参数 + +首先我们dfs函数一定要存一个图，用来遍历的，还要存一个目前我们遍历的节点，定义为x + +至于 单一路径，和路径集合可以放在全局变量，那么代码是这样的： + +```c++ +vector> result; // 收集符合条件的路径 +vector path; // 0节点到终点的路径 +// x：目前遍历的节点 +// graph：存当前的图 +void dfs (vector>& graph, int x) +``` + +2. 确认终止条件 + +什么时候我们就找到一条路径了？ + +当目前遍历的节点 为 最后一个节点的时候，就找到了一条，从 出发点到终止点的路径。 + +当前遍历的节点，我们定义为x，最后一点节点，就是 graph.size() - 1。 + +所以 但 x 等于 graph.size() - 1 的时候就找到一条有效路径。 代码如下： + + +```c++ +// 要求从节点 0 到节点 n-1 的路径并输出，所以是 graph.size() - 1 +if (x == graph.size() - 1) { // 找到符合条件的一条路径 + result.push_back(path); // 收集有效路径 + return; +} +``` + +3. 处理目前搜索节点出发的路径 + +接下来是走 当前遍历节点x的下一个节点。 + +首先是要找到 x节点链接了哪些节点呢？ 遍历方式是这样的： + +```c++ +for (int i = 0; i < graph[x].size(); i++) { // 遍历节点n链接的所有节点 +``` + +接下来就是将 选中的x所连接的节点，加入到 单一路劲来。 + +```C++ +path.push_back(graph[x][i]); // 遍历到的节点加入到路径中来 + +``` + +当前遍历的节点就是 `graph[x][i]` 了，所以进入下一层递归 + +```C++ +dfs(graph, graph[x][i]); // 进入下一层递归 +``` + +最后就是回溯的过程，撤销本次添加节点的操作。 该过程整体代码： + +```C++ +for (int i = 0; i < graph[x].size(); i++) { // 遍历节点n链接的所有节点 + path.push_back(graph[x][i]); // 遍历到的节点加入到路径中来 + dfs(graph, graph[x][i]); // 进入下一层递归 + path.pop_back(); // 回溯，撤销本节点 +} +``` + + +### 本题代码 + +```c++ +class Solution { +private: + vector> result; // 收集符合条件的路径 + vector path; // 0节点到终点的路径 + // x：目前遍历的节点 + // graph：存当前的图 + void dfs (vector>& graph, int x) { + // 要求从节点 0 到节点 n-1 的路径并输出，所以是 graph.size() - 1 + if (x == graph.size() - 1) { // 找到符合条件的一条路径 + result.push_back(path); + return; + } + for (int i = 0; i < graph[x].size(); i++) { // 遍历节点n链接的所有节点 + path.push_back(graph[x][i]); // 遍历到的节点加入到路径中来 + dfs(graph, graph[x][i]); // 进入下一层递归 + path.pop_back(); // 回溯，撤销本节点 + } + } +public: + vector> allPathsSourceTarget(vector>& graph) { + path.push_back(0); // 无论什么路径已经是从0节点出发 + dfs(graph, 0); // 开始遍历 + return result; + } +}; + +``` + +## 其他语言版本 + +### Java + +### Python + +### Go diff --git a/problems/0841.钥匙和房间.md b/problems/0841.钥匙和房间.md index 6f51b4ad..58765a8f 100644 --- a/problems/0841.钥匙和房间.md +++ b/problems/0841.钥匙和房间.md @@ -31,21 +31,182 @@ * 解释：我们不能进入 2 号房间。 -## 思 +## 思路 -其实这道题的本质就是判断各个房间所连成的有向图，说明不用访问所有的房间。 +本题其实给我们是一个有向图， 意识到这是有向图很重要！ -如图所示： +图中给我的两个示例： `[[1],[2],[3],[]]` `[[1,3],[3,0,1],[2],[0]]`，画成对应的图如下： - +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20220714101414.png) -示例1就可以访问所有的房间，因为通过房间里的key将房间连在了一起。 +我们可以看出图1的所有节点都是链接的，而图二中，节点2 是孤立的。 -示例2中，就不能访问所有房间，从图中就可以看出，房间2是一个孤岛，我们从0出发，无论怎么遍历，都访问不到房间2。 +这就很容易让我们想起岛屿问题，只要发现独立的岛，就是不能进入所有房间。 -认清本质问题之后，**使用 广度优先搜索(BFS) 还是 深度优先搜索(DFS) 都是可以的。** +此时也容易想到用并查集的方式去解决。 -BFS C++代码代码如下： +**但本题是有向图**，在有向图中，即使所有节点都是链接的，但依然不可能从0出发遍历所有边。 +给大家举一个例子： + +图3：[[5], [], [1, 3], [5]] ，如图： + +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20220714102201.png) + +在图3中，大家可以发现，节点0只能到节点5，然后就哪也去不了了。 + +所以本题是一个有向图搜索全路径的问题。 只能用深搜（BFS）或者广搜（DFS）来搜。 + +关于DFS的理论，如果大家有困惑，可以先看我这篇题解： [DFS理论基础](https://leetcode.cn/problems/all-paths-from-source-to-target/solution/by-carlsun-2-66pf) + +**以下dfs分析 大家一定要仔细看，本题有两种dfs的解法，很多题解没有讲清楚**。 看完之后 相信你对dfs会有更深的理解。 + +深搜三部曲： + +1. 确认递归函数，参数 + +需要传入二维数组rooms来遍历地图，需要知道当前我们拿到的key，以至于去下一个房间。 + +同时还需要一个数组，用来记录我们都走过了哪些房间，这样好知道最后有没有把所有房间都遍历的，可以定义一个一维数组。 + +所以 递归函数参数如下： + +```C++ +// key 当前得到的可以 +// visited 记录访问过的房间 +void dfs(const vector>& rooms, int key, vector& visited) { +``` + +2. 确认终止条件 + +遍历的时候，什么时候终止呢？ + +这里有一个很重要的逻辑，就是在递归中，**我们是处理当前访问的节点，还是处理下一个要访问的节点**。 + +这决定 终止条件怎么写。 + +首先明确，本题中什么叫做处理，就是 visited数组来记录访问过的节点，那么把该节点默认 数组里元素都是false，把元素标记为true就是处理 本节点了。 + +如果我们是处理当前访问的节点，当前访问的节点如果是 true ，说明是访问过的节点，那就终止本层递归，如果不是true，我们就把它赋值为true，因为我们处理本层递归的节点。 + +代码就是这样： + +```C++ +// 写法一：处理当前访问的节点 +void dfs(const vector>& rooms, int key, vector& visited) { + if (visited[key]) { // 本层递归是true，说明访问过，立刻返回 + return; + } + visited[key] = true; // 给当前遍历的节点赋值true + vector keys = rooms[key]; + for (int key : keys) { + // 深度优先搜索遍历 + dfs(rooms, key, visited); + } +} +``` + +如果我们是处理下一层访问的节点，而不是当前层。那么就要在 深搜三部曲中第三步：处理目前搜索节点出发的路径 的时候对 节点进行处理。 + +这样的话，就不需要终止条件，而是在 搜索下一个节点的时候，直接判断 下一个节点是否是我们要搜的节点。 + +代码就是这样的： + +```C++ +// 写法二：处理下一个要访问的节点 +void dfs(const vector>& rooms, int key, vector& visited) { + // 这里 没有终止条件，而是在 处理下一层节点的时候来判断 + vector keys = rooms[key]; + for (int key : keys) { + if (visited[key] == false) { // 处理下一层节点，判断是否要进行递归 + visited[key] = true; + dfs(rooms, key, visited); + } + } +} +``` + +可以看出，如果看待 我们要访问的节点，直接决定了两种不一样的写法，很多同学对这一块很模糊，其实做过这道题，也没有思考到这个维度上。 + + +3. 处理目前搜索节点出发的路径 + +其实在上面，深搜三部曲 第二部，就已经讲了，因为终止条件的两种写法， 直接决定了两种不一样的递归写法。 + +这里还有细节： + +看上面两个版本的写法中， 好像没有发现回溯的逻辑。 + +我们都知道，有递归就有回溯，回溯就在递归函数的下面， 那么之前我们做的dfs题目，都需要回溯操作，例如：[797.所有可能的路径](https://leetcode.cn/problems/all-paths-from-source-to-target/solution/by-carlsun-2-66pf/)， **为什么本题就没有回溯呢？** + +代码中可以看到dfs函数下面并没有回溯的操作。 + +此时就要在思考本题的要求了，本题是需要判断 0节点是否能到所有节点，那么我们就没有必要回溯去撤销操作了，只要遍历过的节点一律都标记上。 + +**那什么时候需要回溯操作呢？** + +当我们需要搜索一条可行路径的时候，就需要回溯操作了，因为没有回溯，就没法“调头”， 如果不理解的话，去看我写的 [797.所有可能的路径](https://leetcode.cn/problems/all-paths-from-source-to-target/solution/by-carlsun-2-66pf/) 的题解。 + + +以上分析完毕，DFS整体实现C++代码如下： + +```CPP +// 写法一：处理当前访问的节点 +class Solution { +private: + void dfs(const vector>& rooms, int key, vector& visited) { + if (visited[key]) { + return; + } + visited[key] = true; + vector keys = rooms[key]; + for (int key : keys) { + // 深度优先搜索遍历 + dfs(rooms, key, visited); + } + } +public: + bool canVisitAllRooms(vector>& rooms) { + vector visited(rooms.size(), false); + dfs(rooms, 0, visited); + //检查是否都访问到了 + for (int i : visited) { + if (i == false) return false; + } + return true; + } +}; + +``` + +```c++ +写法二：处理下一个要访问的节点 +class Solution { +private: + void dfs(const vector>& rooms, int key, vector& visited) { + vector keys = rooms[key]; + for (int key : keys) { + if (visited[key] == false) { + visited[key] = true; + dfs(rooms, key, visited); + } + } + } +public: + bool canVisitAllRooms(vector>& rooms) { + vector visited(rooms.size(), false); + visited[0] = true; // 0 节点是出发节点，一定被访问过 + dfs(rooms, 0, visited); + //检查是否都访问到了 + for (int i : visited) { + if (i == false) return false; + } + return true; + } +}; + +``` + +本题我也给出 BFS C++代码，至于BFS，我后面会有单独文章来讲，代码如下： ```CPP class Solution { @@ -80,39 +241,11 @@ public: }; ``` -DFS C++代码如下： - -```CPP -class Solution { -private: - void dfs(int key, const vector>& rooms, vector& visited) { - if (visited[key]) { - return; - } - visited[key] = 1; - vector keys = rooms[key]; - for (int key : keys) { - // 深度优先搜索遍历 - dfs(key, rooms, visited); - } - } -public: - bool canVisitAllRooms(vector>& rooms) { - vector visited(rooms.size(), 0); - dfs(0, rooms, visited); - //检查是否都访问到了 - for (int i : visited) { - if (i == 0) return false; - } - return true; - } -}; -``` -# 其他语言版本 +## 其他语言版本 -Java： +### Java ```java class Solution { @@ -120,24 +253,19 @@ class Solution { if (visited.get(key)) { return; } - visited.set(key, true); for (int k : rooms.get(key)) { // 深度优先搜索遍历 dfs(k, rooms, visited); } } - - public boolean canVisitAllRooms(List> rooms) { List visited = new ArrayList(){{ for(int i = 0 ; i < rooms.size(); i++){ add(false); } }}; - dfs(0, rooms, visited); - //检查是否都访问到了 for (boolean flag : visited) { if (!flag) { @@ -149,20 +277,14 @@ class Solution { } ``` - - - - -python3 +### python3 ```python class Solution: - def dfs(self, key: int, rooms: List[List[int]] , visited : List[bool] ) : if visited[key] : return - visited[key] = True keys = rooms[key] for i in range(len(keys)) : @@ -183,7 +305,7 @@ class Solution: ``` -Go： +### Go ```go @@ -201,11 +323,8 @@ func dfs(key int, rooms [][]int, visited []bool ) { } func canVisitAllRooms(rooms [][]int) bool { - visited := make([]bool, len(rooms)); - dfs(0, rooms, visited); - //检查是否都访问到了 for i := 0; i < len(visited); i++ { if !visited[i] { @@ -216,7 +335,7 @@ func canVisitAllRooms(rooms [][]int) bool { } ``` -JavaScript： +### JavaScript ```javascript //DFS var canVisitAllRooms = function(rooms) { diff --git a/problems/1791.找出星型图的中心节点.md b/problems/1791.找出星型图的中心节点.md new file mode 100644 index 00000000..57777fd7 --- /dev/null +++ b/problems/1791.找出星型图的中心节点.md @@ -0,0 +1,73 @@ +# 1791.找出星型图的中心节点 + +[题目链接](https://leetcode.cn/problems/find-center-of-star-graph/) + +本题思路就是统计各个节点的度（这里没有区别入度和出度），如果某个节点的度等于这个图边的数量。 那么这个节点一定是中心节点。 + +什么是度，可以理解为，链接节点的边的数量。 题目中度如图所示： + +![1791.找出星型图的中心节点](https://code-thinking-1253855093.file.myqcloud.com/pics/20220704113207.png) + +至于出度和入度，那就是在有向图里的概念了，本题是无向图。 + +本题代码如下： + +```c++ + +class Solution { +public: + int findCenter(vector>& edges) { + unordered_map du; + for (int i = 0; i < edges.size(); i++) { // 统计各个节点的度 + du[edges[i][1]]++; + du[edges[i][0]]++; + } + unordered_map::iterator iter; // 找出度等于边熟练的节点 + for (iter = du.begin(); iter != du.end(); iter++) { + if (iter->second == edges.size()) return iter->first; + } + return -1; + } +}; +``` + +其实可以只记录度不用最后统计，因为题目说了一定是星状图，所以 一旦有 节点的度 大于1，就返回该节点数值就行，只有中心节点的度会大于1。 + +代码如下： + +```c++ +class Solution { +public: + int findCenter(vector>& edges) { + vector du(edges.size() + 2); // edges.size() + 1 为节点数量，下标表示节点数，所以+2 + for (int i = 0; i < edges.size(); i++) { + du[edges[i][1]]++; + du[edges[i][0]]++; + if (du[edges[i][1]] > 1) return edges[i][1]; + if (du[edges[i][0]] > 1) return edges[i][0]; + + } + return -1; + } +}; +``` + +以上代码中没有使用 unordered_map，因为遍历的时候，开辟新空间会浪费时间，而采用 vector，这是 空间换时间的一种策略。 + +代码其实可以再精简： + +```c++ +class Solution { +public: + int findCenter(vector>& edges) { + vector du(edges.size() + 2); + for (int i = 0; i < edges.size(); i++) { + if (++du[edges[i][1]] > 1) return edges[i][1]; + if (++du[edges[i][0]] > 1) return edges[i][0]; + } + return -1; + } +}; +``` + + diff --git a/problems/1971.寻找图中是否存在路径.md b/problems/1971.寻找图中是否存在路径.md new file mode 100644 index 00000000..bf1e93cf --- /dev/null +++ b/problems/1971.寻找图中是否存在路径.md @@ -0,0 +1,123 @@ +# 1971. 寻找图中是否存在路径 + +[题目链接](https://leetcode.cn/problems/find-if-path-exists-in-graph/) + +有一个具有 n个顶点的 双向 图，其中每个顶点标记从 0 到 n - 1（包含 0 和 n - 1）。图中的边用一个二维整数数组 edges 表示，其中 edges[i] = [ui, vi] 表示顶点 ui 和顶点 vi 之间的双向边。 每个顶点对由 最多一条 边连接，并且没有顶点存在与自身相连的边。 + +请你确定是否存在从顶点 start 开始，到顶点 end 结束的 有效路径 。 + +给你数组 edges 和整数 n、start和end，如果从 start 到 end 存在 有效路径 ，则返回 true，否则返回 false 。 + + +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20220705101442.png) + + +提示: + +* 1 <= n <= 2 * 10^5 +* 0 <= edges.length <= 2 * 10^5 +* edges[i].length == 2 +* 0 <= ui, vi <= n - 1 +* ui != vi +* 0 <= start, end <= n - 1 +* 不存在双向边 +* 不存在指向顶点自身的边 + +## 思路 + +这道题目也是并查集基础题目。 + +首先要知道并查集可以解决什么问题呢？ + +主要就是集合问题，两个节点在不在一个集合，也可以将两个节点添加到一个集合中。 + +这里整理出我的并查集模板如下： + +```CPP +int n = 1005; // 节点数量3 到 1000 +int father[1005]; + +// 并查集初始化 +void init() { + for (int i = 0; i < n; ++i) { + father[i] = i; + } +} +// 并查集里寻根的过程 +int find(int u) { + return u == father[u] ? u : father[u] = find(father[u]); +} +// 将v->u 这条边加入并查集 +void join(int u, int v) { + u = find(u); + v = find(v); + if (u == v) return ; + father[v] = u; +} +// 判断 u 和 v是否找到同一个根 +bool same(int u, int v) { + u = find(u); + v = find(v); + return u == v; +} +``` + +以上模板汇总，只要修改 n 和father数组的大小就可以了。 + +并查集主要有三个功能。 + +1. 寻找根节点，函数：find(int u)，也就是判断这个节点的祖先节点是哪个 +2. 将两个节点接入到同一个集合，函数：join(int u, int v)，将两个节点连在同一个根节点上 +3. 判断两个节点是否在同一个集合，函数：same(int u, int v)，就是判断两个节点是不是同一个根节点 + +简单介绍并查集之后，我们再来看一下这道题目。 + +为什么说这道题目是并查集基础题目，因为 可以直接套用模板。 + +使用join(int u, int v)将每条边加入到并查集。 + +最后 same(int u, int v) 判断是否是同一个根 就可以里。 + +代码如下： + +```c++ +class Solution { + +private: + int n = 200005; // 节点数量 20000 + int father[200005]; + + // 并查集初始化 + void init() { + for (int i = 0; i < n; ++i) { + father[i] = i; + } + } + // 并查集里寻根的过程 + int find(int u) { + return u == father[u] ? u : father[u] = find(father[u]); + } + // 将v->u 这条边加入并查集 + void join(int u, int v) { + u = find(u); + v = find(v); + if (u == v) return ; + father[v] = u; + } + // 判断 u 和 v是否找到同一个根，本题用不上 + bool same(int u, int v) { + u = find(u); + v = find(v); + return u == v; + } + +public: + bool validPath(int n, vector>& edges, int source, int destination) { + init(); + for (int i = 0; i < edges.size(); i++) { + join(edges[i][0], edges[i][1]); + } + return same(source, destination); + } +}; +``` diff --git a/problems/动态规划总结篇.md b/problems/动态规划总结篇.md index cc973b23..847baa9a 100644 --- a/problems/动态规划总结篇.md +++ b/problems/动态规划总结篇.md @@ -116,7 +116,7 @@ 能把本篇中列举的题目都研究通透的话，你的动规水平就已经非常高了。 对付面试已经足够！ -![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211121223754.png) +![](https://kstar-1253855093.cos.ap-nanjing.myqcloud.com/baguwenpdf/_%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%E6%80%9D%E7%BB%B4%E5%AF%BC%E5%9B%BE_%E9%9D%92.png) 这个图是 [代码随想录知识星球](https://programmercarl.com/other/kstar.html) 成员：[青](https://wx.zsxq.com/dweb2/index/footprint/185251215558842)，所画，总结的非常好，分享给大家。 diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 1ae01061..c6d33fe1 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -94,8 +94,8 @@ public: }; ``` -* 时间复杂度：$O(n + m)$ -* 空间复杂度：$O(1)$ +* 时间复杂度：O(n + m) +* 空间复杂度：O(1) ## 其他语言版本 From e108c87570953684cf1aee862d580d5d220bcdb4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Thu, 21 Jul 2022 09:37:53 +0800 Subject: [PATCH 129/136] =?UTF-8?q?Update=200494.=E7=9B=AE=E6=A0=87?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0494.目标和.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 60f721c2..b8707581 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -213,7 +213,7 @@ public: if (abs(S) > sum) return 0; // 此时没有方案 if ((S + sum) % 2 == 1) return 0; // 此时没有方案 int bagSize = (S + sum) / 2; - if(bagsize<0) return 0; + if (bagsize < 0) return 0; vector dp(bagSize + 1, 0); dp[0] = 1; for (int i = 0; i < nums.size(); i++) { From d59bc2ee161e1ca76642a72cfabd83f8f4218589 Mon Sep 17 00:00:00 2001 From: cezarbbb <105843128+cezarbbb@users.noreply.github.com> Date: Thu, 21 Jul 2022 10:33:02 +0800 Subject: [PATCH 130/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200046.=E5=85=A8?= =?UTF-8?q?=E6=8E=92=E5=88=97=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 0046.全排列 Rust版本 --- problems/0046.全排列.md | 30 ++++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md index 06e1550a..ce07395a 100644 --- a/problems/0046.全排列.md +++ b/problems/0046.全排列.md @@ -359,6 +359,36 @@ function permute(nums: number[]): number[][] { }; ``` +### Rust + +```Rust +impl Solution { + fn backtracking(result: &mut Vec>, path: &mut Vec, nums: &Vec, used: &mut Vec) { + let len = nums.len(); + if path.len() == len { + result.push(path.clone()); + return; + } + for i in 0..len { + if used[i] == true { continue; } + used[i] = true; + path.push(nums[i]); + Self::backtracking(result, path, nums, used); + path.pop(); + used[i] = false; + } + } + + pub fn permute(nums: Vec) -> Vec> { + let mut result: Vec> = Vec::new(); + let mut path: Vec = Vec::new(); + let mut used = vec![false; nums.len()]; + Self::backtracking(&mut result, &mut path, &nums, &mut used); + result + } +} +``` + ### C ```c From bafb7c45db1b9975c590ec93a65034c7b2190656 Mon Sep 17 00:00:00 2001 From: lizhendong128 Date: Thu, 21 Jul 2022 19:50:02 +0800 Subject: [PATCH 131/136] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200309.=E6=9C=80?= =?UTF-8?q?=E4=BD=B3=E4=B9=B0=E5=8D=96=E8=82=A1=E7=A5=A8=E6=97=B6=E6=9C=BA?= =?UTF-8?q?=E5=90=AB=E5=86=B7=E5=86=BB=E6=9C=9F=20=E7=94=A8=E4=BA=8E?= =?UTF-8?q?=E8=AF=B4=E6=98=8E=E7=9A=84=E5=9B=BE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0309.最佳买卖股票时机含冷冻期.md | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) diff --git a/problems/0309.最佳买卖股票时机含冷冻期.md b/problems/0309.最佳买卖股票时机含冷冻期.md index 229fc636..3791a71f 100644 --- a/problems/0309.最佳买卖股票时机含冷冻期.md +++ b/problems/0309.最佳买卖股票时机含冷冻期.md @@ -44,6 +44,8 @@ dp[i][j]，第i天状态为j，所剩的最多现金为dp[i][j]。 * 状态三：今天卖出了股票 * 状态四：今天为冷冻期状态，但冷冻期状态不可持续，只有一天！ +![](https://img-blog.csdnimg.cn/518d5baaf33f4b2698064f8efb42edbf.png) + j的状态为： * 0：状态一 @@ -57,7 +59,7 @@ j的状态为： **注意这里的每一个状态，例如状态一，是买入股票状态并不是说今天已经就买入股票，而是说保存买入股票的状态即：可能是前几天买入的，之后一直没操作，所以保持买入股票的状态**。 -2. 确定递推公式 +1. 确定递推公式 达到买入股票状态（状态一）即：dp[i][0]，有两个具体操作： From 8865ed851aeb14a7e768c9fa5d7ef0c2a110c56f Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E6=A1=9C=E5=B0=8F=E8=B7=AF=E4=B8=83=E8=91=89?= <20304773@qq.com> Date: Fri, 22 Jul 2022 15:36:23 +0800 Subject: [PATCH 132/136] =?UTF-8?q?Update=200941.=E6=9C=89=E6=95=88?= =?UTF-8?q?=E7=9A=84=E5=B1=B1=E8=84=89=E6=95=B0=E7=BB=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0941.有效的山脉数组.md | 17 +++++++++++++++++ 1 file changed, 17 insertions(+) diff --git a/problems/0941.有效的山脉数组.md b/problems/0941.有效的山脉数组.md index 310dd35a..fb7935a8 100644 --- a/problems/0941.有效的山脉数组.md +++ b/problems/0941.有效的山脉数组.md @@ -177,7 +177,24 @@ function validMountainArray(arr: number[]): boolean { }; ``` +## C# +```csharp +public class Solution { + public bool ValidMountainArray(int[] arr) { + if (arr.Length < 3) return false; + + int left = 0; + int right = arr.Length - 1; + + while (left + 1< arr.Length && arr[left] < arr[left + 1]) left ++; + while (right > 0 && arr[right] < arr[right - 1]) right --; + if (left == right && left != 0 && right != arr.Length - 1) return true; + + return false; + } +} +``` ----------------------- From f5b44d5f8c8ed2136a396a104b4f62884319eede Mon Sep 17 00:00:00 2001 From: Luo <82520819+Jerry-306@users.noreply.github.com> Date: Sat, 23 Jul 2022 13:06:47 +0800 Subject: [PATCH 133/136] =?UTF-8?q?js=20=E4=BB=A3=E7=A0=81=E6=97=A0?= =?UTF-8?q?=E9=AB=98=E4=BA=AE=E6=8F=90=E7=A4=BA?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/回溯算法去重问题的另一种写法.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/回溯算法去重问题的另一种写法.md b/problems/回溯算法去重问题的另一种写法.md index cbfe046a..c7bf24bc 100644 --- a/problems/回溯算法去重问题的另一种写法.md +++ b/problems/回溯算法去重问题的另一种写法.md @@ -418,7 +418,7 @@ function combinationSum2(candidates, target) { **47. 全排列 II** -```javaescript +```javascript function permuteUnique(nums) { const resArr = []; const usedArr = []; From fdf0185e1d53e8cda56a6cb040da5b6218fbb3c3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Tue, 26 Jul 2022 09:17:40 +0800 Subject: [PATCH 134/136] =?UTF-8?q?Update=200541.=E5=8F=8D=E8=BD=AC?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0541.反转字符串II.md | 34 ++++++++++++++++++---------------- 1 file changed, 18 insertions(+), 16 deletions(-) diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md index d9b9466c..8c654a17 100644 --- a/problems/0541.反转字符串II.md +++ b/problems/0541.反转字符串II.md @@ -64,22 +64,7 @@ public: ``` -``` -class Solution { -public: - string reverseStr(string s, int k) { - int n=s.size(),pos=0; - while(pos Date: Sat, 30 Jul 2022 09:45:38 +0800 Subject: [PATCH 135/136] =?UTF-8?q?0452.=E7=94=A8=E6=9C=80=E5=B0=91?= =?UTF-8?q?=E6=95=B0=E9=87=8F=E7=9A=84=E7=AE=AD=E5=BC=95=E7=88=86=E6=B0=94?= =?UTF-8?q?=E7=90=83=20Java=E4=BB=A3=E7=A0=81=E4=BF=AE=E6=94=B9?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0452.用最少数量的箭引爆气球.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md index 3042b063..7b5f387d 100644 --- a/problems/0452.用最少数量的箭引爆气球.md +++ b/problems/0452.用最少数量的箭引爆气球.md @@ -151,6 +151,7 @@ class Solution { //重叠气球的最小右边界 int leftmostRightBound = points[0][1]; //如果下一个气球的左边界大于最小右边界 + for(int i = 1; i < points.length; i++){ if (points[i][0] > leftmostRightBound ) { //增加一次射击 count++; From 9343ddd30cb63ad51db6f023c581f31b1f3f9d9a Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Mon, 1 Aug 2022 10:50:27 +0800 Subject: [PATCH 136/136] =?UTF-8?q?Update=200494.=E7=9B=AE=E6=A0=87?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0494.目标和.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 00e4bdfa..cb1b9bd1 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -158,7 +158,7 @@ dp[j] 表示：填满j（包括j）这么大容积的包，有dp[j]种方法 不考虑nums[i]的情况下，填满容量为j的背包，有dp[j]种方法。 -那么只要搞到nums[i]的话，凑成dp[j]就有dp[j - nums[i]] 种方法。 +那么考虑nums[i]的话（只要搞到nums[i]），凑成dp[j]就有dp[j - nums[i]] 种方法。 例如：dp[j]，j 为5，