diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index 055f8940..a15e1349 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -335,6 +335,20 @@ 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]; +} +``` -----------------------
diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md index 5a4168d1..f336be5b 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。 @@ -411,6 +411,76 @@ var threeSum = function(nums) { return res }; ``` + +解法二：nSum通用解法。递归 + +```js +/** + * nsum通用解法，支持2sum，3sum，4sum...等等 + * 时间复杂度分析： + * 1. n = 2时，时间复杂度O(NlogN)，排序所消耗的时间。、 + * 2. n > 2时，时间复杂度为O(N^n-1)，即N的n-1次方，至少是2次方，此时可省略排序所消耗的时间。举例：3sum为O(n^2)，4sum为O(n^3) + * @param {number[]} nums + * @return {number[][]} + */ +var threeSum = function (nums) { + // nsum通用解法核心方法 + function nSumTarget(nums, n, start, target) { + // 前提：nums要先排序好 + let res = []; + if (n === 2) { + res = towSumTarget(nums, start, target); + } else { + for (let i = start; i < nums.length; i++) { + // 递归求(n - 1)sum + let subRes = nSumTarget( + nums, + n - 1, + i + 1, + target - nums[i] + ); + for (let j = 0; j < subRes.length; j++) { + res.push([nums[i], ...subRes[j]]); + } + // 跳过相同元素 + while (nums[i] === nums[i + 1]) i++; + } + } + return res; + } + + function towSumTarget(nums, start, target) { + // 前提：nums要先排序好 + let res = []; + let len = nums.length; + let left = start; + let right = len - 1; + while (left < right) { + let sum = nums[left] + nums[right]; + if (sum < target) { + while (nums[left] === nums[left + 1]) left++; + left++; + } else if (sum > target) { + while (nums[right] === nums[right - 1]) right--; + right--; + } else { + // 相等 + res.push([nums[left], nums[right]]); + // 跳过相同元素 + while (nums[left] === nums[left + 1]) left++; + while (nums[right] === nums[right - 1]) right--; + left++; + right--; + } + } + return res; + } + nums.sort((a, b) => a - b); + // n = 3，此时求3sum之和 + return nSumTarget(nums, 3, 0, 0); +}; +``` + TypeScript: ```typescript @@ -550,6 +620,71 @@ 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/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 2b5cb978..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 @@ -557,6 +600,37 @@ func letterCombinations(_ digits: String) -> [String] { } ``` +## Scala: + +```scala +object Solution { + import scala.collection.mutable + def letterCombinations(digits: String): List[String] = { + var result = mutable.ListBuffer[String]() + if(digits == "") return result.toList // 如果参数为空，返回空结果集的List形式 + var path = mutable.ListBuffer[Char]() + // 数字和字符的映射关系 + val map = Array[String]("", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz") + + def backtracking(index: Int): Unit = { + if (index == digits.size) { + result.append(path.mkString) // mkString语法：将数组类型直接转换为字符串 + return + } + var digit = digits(index) - '0' // 这里使用toInt会报错！必须 -'0' + for (i <- 0 until map(digit).size) { + path.append(map(digit)(i)) + backtracking(index + 1) + path = path.take(path.size - 1) + } + } + + backtracking(0) + result.toList + } +} +``` + -----------------------
diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 7989ad8f..3813424a 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -135,6 +135,11 @@ 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; } @@ -148,7 +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) { @@ -517,6 +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 { diff --git a/problems/0020.有效的括号.md b/problems/0020.有效的括号.md index b44d21ed..c3ff9d53 100644 --- a/problems/0020.有效的括号.md +++ b/problems/0020.有效的括号.md @@ -400,6 +400,37 @@ bool isValid(char * s){ return !stackTop; } ``` + + +PHP: +```php +// https://www.php.net/manual/zh/class.splstack.php +class Solution +{ + function isValid($s){ + $stack = new SplStack(); + for ($i = 0; $i < strlen($s); $i++) { + if ($s[$i] == "(") { + $stack->push(')'); + } else if ($s[$i] == "{") { + $stack->push('}'); + } else if ($s[$i] == "[") { + $stack->push(']'); + // 2、遍历匹配过程中，发现栈内没有要匹配的字符 return false + // 3、遍历匹配过程中，栈已为空，没有匹配的字符了，说明右括号没有找到对应的左括号 return false + } else if ($stack->isEmpty() || $stack->top() != $s[$i]) { + return false; + } else {//$stack->top() == $s[$i] + $stack->pop(); + } + } + // 1、遍历完，但是栈不为空,说明有相应的括号没有被匹配,return false + return $stack->isEmpty(); + } +} +``` + + Scala: ```scala object Solution { @@ -422,5 +453,6 @@ object Solution { } } ``` + -----------------------
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 550a886e..4b4b39e7 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 @@ -256,20 +256,19 @@ 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: diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 03c58b43..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 } ``` @@ -339,7 +340,6 @@ int removeElement(int* nums, int numsSize, int val){ } ``` - Kotlin: ```kotlin fun removeElement(nums: IntArray, `val`: Int): Int { @@ -351,7 +351,6 @@ fun removeElement(nums: IntArray, `val`: Int): Int { } ``` - Scala: ```scala object Solution { @@ -368,5 +367,20 @@ object Solution { } ``` +C#: +```csharp +public class Solution { + public int RemoveElement(int[] nums, int val) { + int slow = 0; + for (int fast = 0; fast < nums.Length; fast++) { + if (val != nums[fast]) { + nums[slow++] = nums[fast]; + } + } + return slow; + } +} +``` + -----------------------
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/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= target) { + // 左边界一定在mid左边（不含mid） + right = mid - 1; + leftBoard = right; + } else { + // 左边界在mid右边（含mid） + left = mid + 1; + } + } + return leftBoard; +} +``` + + ### Scala ```scala object Solution { @@ -529,5 +585,49 @@ 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/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md index ed7c49b4..77def4f8 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 diff --git a/problems/0037.解数独.md b/problems/0037.解数独.md index b074b98f..626e9d97 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 + } +} +``` + -----------------------
diff --git a/problems/0039.组合总和.md b/problems/0039.组合总和.md index fef9b676..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 ``` @@ -502,5 +502,35 @@ func combinationSum(_ candidates: [Int], _ target: Int) -> [[Int]] { } ``` +## Scala + +```scala +object Solution { + import scala.collection.mutable + def combinationSum(candidates: Array[Int], target: Int): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() + var path = mutable.ListBuffer[Int]() + + def backtracking(sum: Int, index: Int): Unit = { + if (sum == target) { + result.append(path.toList) // 如果正好等于target，就添加到结果集 + return + } + // 应该是从当前索引开始的，而不是从0 + // 剪枝优化：添加循环守卫，当sum + c(i) <= target的时候才循环，才可以进入下一次递归 + for (i <- index until candidates.size if sum + candidates(i) <= target) { + path.append(candidates(i)) + backtracking(sum + candidates(i), i) + path = path.take(path.size - 1) + } + } + + backtracking(0, 0) + result.toList + } +} +``` + + -----------------------
diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md index ee4371cd..557eb855 100644 --- a/problems/0040.组合总和II.md +++ b/problems/0040.组合总和II.md @@ -693,5 +693,37 @@ func combinationSum2(_ candidates: [Int], _ target: Int) -> [[Int]] { } ``` + +## Scala + +```scala +object Solution { + import scala.collection.mutable + def combinationSum2(candidates: Array[Int], target: Int): List[List[Int]] = { + var res = mutable.ListBuffer[List[Int]]() + var path = mutable.ListBuffer[Int]() + var candidate = candidates.sorted + + def backtracking(sum: Int, startIndex: Int): Unit = { + if (sum == target) { + res.append(path.toList) + return + } + + for (i <- startIndex until candidate.size if sum + candidate(i) <= target) { + if (!(i > startIndex && candidate(i) == candidate(i - 1))) { + path.append(candidate(i)) + backtracking(sum + candidate(i), i + 1) + path = path.take(path.size - 1) + } + } + } + + backtracking(0, 0) + res.toList + } +} +``` + -----------------------
diff --git a/problems/0042.接雨水.md b/problems/0042.接雨水.md index 4832423d..448d6d51 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 diff --git a/problems/0045.跳跃游戏II.md b/problems/0045.跳跃游戏II.md index 9e61fbb8..648355fd 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 + } +} +``` diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md index 9f93564c..06e1550a 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 + } +} +``` -----------------------
diff --git a/problems/0047.全排列II.md b/problems/0047.全排列II.md index f5f9e440..88680c8c 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 + } +} +``` + -----------------------
diff --git a/problems/0051.N皇后.md b/problems/0051.N皇后.md index 798902ae..9ae1f975 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 + } +} +``` + -----------------------
diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index cbc91481..17b9d31e 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 + } +} +``` -----------------------
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 345abc27..228ceade 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 @@ -186,6 +186,24 @@ 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 + } +} +``` + TypeScript： ```typescript @@ -205,6 +223,5 @@ function maxSubArray(nums: number[]): number { ``` - -----------------------
diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index 6fa83495..e7c4f5a2 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 + } +} +``` ----------------------- diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index 757896c1..a8f36fdf 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -133,24 +133,38 @@ 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 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 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 diff --git a/problems/0077.组合.md b/problems/0077.组合.md index 8d22d018..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; @@ -673,5 +723,63 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] { } ``` +### Scala + +暴力: +```scala +object Solution { + import scala.collection.mutable // 导包 + def combine(n: Int, k: Int): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() // 存放结果集 + var path = mutable.ListBuffer[Int]() //存放符合条件的结果 + + def backtracking(n: Int, k: Int, startIndex: Int): Unit = { + if (path.size == k) { + // 如果path的size == k就达到题目要求，添加到结果集，并返回 + result.append(path.toList) + return + } + for (i <- startIndex to n) { // 遍历从startIndex到n + path.append(i) // 先把数字添加进去 + backtracking(n, k, i + 1) // 进行下一步回溯 + path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字 + } + } + + backtracking(n, k, 1) // 执行回溯 + result.toList // 最终返回result的List形式，return关键字可以省略 + } +} +``` + +剪枝: + +```scala +object Solution { + import scala.collection.mutable // 导包 + def combine(n: Int, k: Int): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() // 存放结果集 + var path = mutable.ListBuffer[Int]() //存放符合条件的结果 + + def backtracking(n: Int, k: Int, startIndex: Int): Unit = { + if (path.size == k) { + // 如果path的size == k就达到题目要求，添加到结果集，并返回 + result.append(path.toList) + return + } + // 剪枝优化 + for (i <- startIndex to (n - (k - path.size) + 1)) { + path.append(i) // 先把数字添加进去 + backtracking(n, k, i + 1) // 进行下一步回溯 + path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字 + } + } + + backtracking(n, k, 1) // 执行回溯 + result.toList // 最终返回result的List形式，return关键字可以省略 + } +} +``` + -----------------------
diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md index a6767047..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 @@ -346,5 +372,34 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] { } ``` +Scala: + +```scala +object Solution { + import scala.collection.mutable // 导包 + def combine(n: Int, k: Int): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() // 存放结果集 + var path = mutable.ListBuffer[Int]() //存放符合条件的结果 + + def backtracking(n: Int, k: Int, startIndex: Int): Unit = { + if (path.size == k) { + // 如果path的size == k就达到题目要求，添加到结果集，并返回 + result.append(path.toList) + return + } + // 剪枝优化 + for (i <- startIndex to (n - (k - path.size) + 1)) { + path.append(i) // 先把数字添加进去 + backtracking(n, k, i + 1) // 进行下一步回溯 + path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字 + } + } + + backtracking(n, k, 1) // 执行回溯 + result.toList // 最终返回result的List形式，return关键字可以省略 + } +} +``` + -----------------------
diff --git a/problems/0078.子集.md b/problems/0078.子集.md index 3d850c08..e6cc668b 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 + } +} +``` + -----------------------
diff --git a/problems/0090.子集II.md b/problems/0090.子集II.md index 909c5692..e85ec66d 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 + } +} +``` -----------------------
diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md index 41940487..11ca2d03 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 + } +} +``` + -----------------------
diff --git a/problems/0098.验证二叉搜索树.md b/problems/0098.验证二叉搜索树.md index 61dd5427..cba450e5 100644 --- a/problems/0098.验证二叉搜索树.md +++ b/problems/0098.验证二叉搜索树.md @@ -589,7 +589,50 @@ function isValidBST(root: TreeNode | null): boolean { }; ``` +## Scala +辅助数组解决: +```scala +object Solution { + import scala.collection.mutable + def isValidBST(root: TreeNode): Boolean = { + var arr = new mutable.ArrayBuffer[Int]() + // 递归中序遍历二叉树，将节点添加到arr + def traversal(node: TreeNode): Unit = { + if (node == null) return + traversal(node.left) + arr.append(node.value) + traversal(node.right) + } + traversal(root) + // 这个数组如果是升序就代表是二叉搜索树 + for (i <- 1 until arr.size) { + if (arr(i) <= arr(i - 1)) return false + } + true + } +} +``` + +递归中解决: +```scala +object Solution { + def isValidBST(root: TreeNode): Boolean = { + var flag = true + var preValue:Long = Long.MinValue // 这里要使用Long类型 + + def traversal(node: TreeNode): Unit = { + if (node == null || flag == false) return + traversal(node.left) + if (node.value > preValue) preValue = node.value + else flag = false + traversal(node.right) + } + traversal(root) + flag + } +} +``` -----------------------
diff --git a/problems/0100.相同的树.md b/problems/0100.相同的树.md index 5288779b..210418f7 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; +}; +``` + + + -----------------------
diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index 40249bb7..fd2d1987 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 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 a2717d09..5ef23944 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 @@ -2629,21 +2629,21 @@ 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 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); } } } 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 6d87c756..91e0c8d8 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; } } diff --git a/problems/0108.将有序数组转换为二叉搜索树.md b/problems/0108.将有序数组转换为二叉搜索树.md index c9c1a693..b5f322f0 100644 --- a/problems/0108.将有序数组转换为二叉搜索树.md +++ b/problems/0108.将有序数组转换为二叉搜索树.md @@ -448,5 +448,27 @@ struct TreeNode* sortedArrayToBST(int* nums, int numsSize) { } ``` +## Scala + +递归: + +```scala +object Solution { + def sortedArrayToBST(nums: Array[Int]): TreeNode = { + def buildTree(left: Int, right: Int): TreeNode = { + if (left > right) return null // 当left大于right的时候，返回空 + // 最中间的节点是当前节点 + var mid = left + (right - left) / 2 + var curNode = new TreeNode(nums(mid)) + curNode.left = buildTree(left, mid - 1) + curNode.right = buildTree(mid + 1, right) + curNode + } + buildTree(0, nums.size - 1) + } +} +``` + + -----------------------
diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 3aa815ab..5bdd4f1b 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 ``` 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 ed8ce592..303108be 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; +}; +``` + + + -----------------------
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 ae9744d1..e4742cf0 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 + } +} +``` + -----------------------
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 ea3845b7..92a72fe3 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/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) { @@ -289,7 +315,40 @@ 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; diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md index a371864d..64d45853 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 + } +} +``` + + -----------------------
diff --git a/problems/0132.分割回文串II.md b/problems/0132.分割回文串II.md index 92f9b602..36db56cb 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 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; +}; +``` + + + -----------------------
diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md index 254f9db0..0b9059dc 100644 --- a/problems/0142.环形链表II.md +++ b/problems/0142.环形链表II.md @@ -303,13 +303,13 @@ 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; } 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 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 05916135..1a90265a 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 @@ -325,6 +325,33 @@ func evalRPN(_ tokens: [String]) -> Int { return stack.last! } ``` + + +PHP： +```php +class Solution { + function evalRPN($tokens) { + $st = new SplStack(); + for($i = 0;$ipush($tokens[$i]); + }else{ + // 是符号进行运算 + $num1 = $st->pop(); + $num2 = $st->pop(); + if ($tokens[$i] == "+") $st->push($num2 + $num1); + if ($tokens[$i] == "-") $st->push($num2 - $num1); + if ($tokens[$i] == "*") $st->push($num2 * $num1); + // 注意处理小数部分 + if ($tokens[$i] == "/") $st->push(intval($num2 / $num1)); + } + } + return $st->pop(); + } +} +``` + Scala: ```scala object Solution { @@ -351,6 +378,7 @@ object Solution { // 最后返回栈顶，不需要加return关键字 stack.pop() } + } ``` ----------------------- diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index 38372f91..d0719469 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -79,7 +79,7 @@ void removeExtraSpaces(string& s) { 逻辑很简单，从前向后遍历，遇到空格了就erase。 -如果不仔细琢磨一下erase的时间复杂读，还以为以上的代码是O(n)的时间复杂度呢。 +如果不仔细琢磨一下erase的时间复杂度，还以为以上的代码是O(n)的时间复杂度呢。 想一下真正的时间复杂度是多少，一个erase本来就是O(n)的操作，erase实现原理题目：[数组：就移除个元素很难么？](https://programmercarl.com/0027.移除元素.html)，最优的算法来移除元素也要O(n)。 @@ -864,7 +864,58 @@ function reverseString(&$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::() + } +} +``` -----------------------
diff --git a/problems/0189.旋转数组.md b/problems/0189.旋转数组.md index 3ffed877..23092f9c 100644 --- a/problems/0189.旋转数组.md +++ b/problems/0189.旋转数组.md @@ -7,6 +7,8 @@ # 189. 旋转数组 +[力扣题目链接](https://leetcode.cn/problems/rotate-array/) + 给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。 进阶： @@ -160,6 +162,27 @@ var rotate = function (nums, k) { }; ``` +## TypeScript + +```typescript +function rotate(nums: number[], k: number): void { + const length: number = nums.length; + k %= length; + reverseByRange(nums, 0, length - 1); + reverseByRange(nums, 0, k - 1); + reverseByRange(nums, k, length - 1); +}; +function reverseByRange(nums: number[], left: number, right: number): void { + while (left < right) { + const temp = nums[left]; + nums[left] = nums[right]; + nums[right] = temp; + left++; + right--; + } +} +``` + ----------------------- diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 0054582c..d14ee770 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 { diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index b79d29a5..8af8675b 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; }; ``` @@ -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 } } ``` @@ -532,5 +534,39 @@ 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/0205.同构字符串.md b/problems/0205.同构字符串.md index 284e7fd9..43e2b0f0 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; +}; +``` + + + -----------------------
diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index c412e840..a6f4382a 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -422,6 +422,41 @@ 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 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 b374aaec..c75df661 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: @@ -465,6 +497,27 @@ 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; + } +} +``` -----------------------
diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md index a8a6567d..1ef278ff 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 @@ -502,5 +531,35 @@ func combinationSum3(_ count: Int, _ targetSum: Int) -> [[Int]] { } ``` +## Scala + +```scala +object Solution { + import scala.collection.mutable + def combinationSum3(k: Int, n: Int): List[List[Int]] = { + var result = mutable.ListBuffer[List[Int]]() + var path = mutable.ListBuffer[Int]() + + def backtracking(k: Int, n: Int, sum: Int, startIndex: Int): Unit = { + if (sum > n) return // 剪枝，如果sum>目标和，就返回 + if (sum == n && path.size == k) { + result.append(path.toList) + return + } + // 剪枝 + for (i <- startIndex to (9 - (k - path.size) + 1)) { + path.append(i) + backtracking(k, n, sum + i, i + 1) + path = path.take(path.size - 1) + } + } + + backtracking(k, n, 0, 1) // 调用递归方法 + result.toList // 最终返回结果集的List形式 + } +} +``` + + -----------------------
diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index 994d1875..40415d8d 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -816,7 +816,6 @@ class MyStack { } ``` Scala: - 使用两个队列模拟栈: ```scala import scala.collection.mutable @@ -897,6 +896,86 @@ class MyStack() { def empty(): Boolean = { queue.isEmpty } + } +``` + + +PHP +> 双对列 +```php +// SplQueue 类通过使用一个双向链表来提供队列的主要功能。(PHP 5 >= 5.3.0, PHP 7, PHP 8) +// https://www.php.net/manual/zh/class.splqueue.php +class MyStack { + public $queueMain; // 保存数据 + public $queueTmp; // 辅助作用 + + function __construct() { + $this->queueMain=new SplQueue(); + $this->queueTmp=new SplQueue(); + } + + // queueMain: 1,2,3 <= add + function push($x) { + $this->queueMain->enqueue($x); + } + + function pop() { + $qmSize = $this->queueMain->Count(); + $qmSize --; + // queueMain: 3,2,1 => pop =>2,1 => add => 2,1 :queueTmp + while($qmSize --){ + $this->queueTmp->enqueue($this->queueMain->dequeue()); + } + // queueMain: 3 + $val = $this->queueMain->dequeue(); + // queueMain <= queueTmp + $this->queueMain = $this->queueTmp; + // 清空queueTmp,下次使用 + $this->queueTmp = new SplQueue(); + return $val; + } + + function top() { + // 底层是双链表实现：从双链表的末尾查看节点 + return $this->queueMain->top(); + } + + function empty() { + return $this->queueMain->isEmpty(); + } +} +``` +> 单对列 +```php +class MyStack { + public $queue; + + function __construct() { + $this->queue=new SplQueue(); + } + + function push($x) { + $this->queue->enqueue($x); + } + + function pop() { + $qmSize = $this->queue->Count(); + $qmSize --; + //queue: 3,2,1 => pop =>2,1 => add => 2,1,3 :queue + while($qmSize --){ + $this->queue->enqueue($this->queue->dequeue()); + } + $val = $this->queue->dequeue(); + return $val; + } + + function top() { + return $this->queue->top(); + } + + function empty() { + return $this->queue->isEmpty(); + } } ``` ----------------------- 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; }; ``` diff --git a/problems/0232.用栈实现队列.md b/problems/0232.用栈实现队列.md index 47993fa1..4662e2f2 100644 --- a/problems/0232.用栈实现队列.md +++ b/problems/0232.用栈实现队列.md @@ -495,6 +495,53 @@ void myQueueFree(MyQueue* obj) { obj->stackOutTop = 0; } ``` + + +PHP: +```php +// SplStack 类通过使用一个双向链表来提供栈的主要功能。[PHP 5 >= 5.3.0, PHP 7, PHP 8] +// https://www.php.net/manual/zh/class.splstack.php +class MyQueue { + // 双栈模拟队列：In栈存储数据；Out栈辅助处理 + private $stackIn; + private $stackOut; + + function __construct() { + $this->stackIn = new SplStack(); + $this->stackOut = new SplStack(); + } + + // In: 1 2 3 <= push + function push($x) { + $this->stackIn->push($x); + } + + function pop() { + $this->peek(); + return $this->stackOut->pop(); + } + + function peek() { + if($this->stackOut->isEmpty()){ + $this->shift(); + } + return $this->stackOut->top(); + } + + function empty() { + return $this->stackOut->isEmpty() && $this->stackIn->isEmpty(); + } + + // 如果Out栈为空，把In栈数据压入Out栈 + // In: 1 2 3 => pop push => 1 2 3 :Out + private function shift(){ + while(!$this->stackIn->isEmpty()){ + $this->stackOut->push($this->stackIn->pop()); + } + } + } +``` + Scala: ```scala class MyQueue() { @@ -533,6 +580,7 @@ class MyQueue() { def empty(): Boolean = { stackIn.isEmpty && stackOut.isEmpty } + } ``` ----------------------- diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md index f591fcef..eeee6fa5 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; +} +``` + ----------------------- diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md index 9ff7e293..ee86d02f 100644 --- a/problems/0235.二叉搜索树的最近公共祖先.md +++ b/problems/0235.二叉搜索树的最近公共祖先.md @@ -381,7 +381,36 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree }; ``` +## Scala +递归: + +```scala +object Solution { + def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = { + // scala中每个关键字都有其返回值，于是可以不写return + if (root.value > p.value && root.value > q.value) lowestCommonAncestor(root.left, p, q) + else if (root.value < p.value && root.value < q.value) lowestCommonAncestor(root.right, p, q) + else root + } +} +``` + +迭代: + +```scala +object Solution { + def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = { + var curNode = root // 因为root是不可变量，所以要赋值给curNode一个可变量 + while(curNode != null){ + if(curNode.value > p.value && curNode.value > q.value) curNode = curNode.left + else if(curNode.value < p.value && curNode.value < q.value) curNode = curNode.right + else return curNode + } + null + } +} +``` ----------------------- diff --git a/problems/0236.二叉树的最近公共祖先.md b/problems/0236.二叉树的最近公共祖先.md index 23695b11..c3e2ae7a 100644 --- a/problems/0236.二叉树的最近公共祖先.md +++ b/problems/0236.二叉树的最近公共祖先.md @@ -343,7 +343,25 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree }; ``` +## Scala +```scala +object Solution { + def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = { + // 递归结束条件 + if (root == null || root == p || root == q) { + return root + } + + var left = lowestCommonAncestor(root.left, p, q) + var right = lowestCommonAncestor(root.right, p, q) + + if (left != null && right != null) return root + if (left == null) return right + left + } +} +``` -----------------------
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 23e79c80..7ee1fdb1 100644 --- a/problems/0239.滑动窗口最大值.md +++ b/problems/0239.滑动窗口最大值.md @@ -654,8 +654,7 @@ object Solution { // 最终返回res，return关键字可以省略 res } - -} + } class MyQueue { var queue = ArrayBuffer[Int]() @@ -678,5 +677,84 @@ class MyQueue { def peek(): Int = queue.head } ``` + + +PHP: +```php +class Solution { + /** + * @param Integer[] $nums + * @param Integer $k + * @return Integer[] + */ + function maxSlidingWindow($nums, $k) { + $myQueue = new MyQueue(); + // 先将前k的元素放进队列 + for ($i = 0; $i < $k; $i++) { + $myQueue->push($nums[$i]); + } + + $result = []; + $result[] = $myQueue->max(); // result 记录前k的元素的最大值 + + for ($i = $k; $i < count($nums); $i++) { + $myQueue->pop($nums[$i - $k]); // 滑动窗口移除最前面元素 + $myQueue->push($nums[$i]); // 滑动窗口前加入最后面的元素 + $result[]= $myQueue->max(); // 记录对应的最大值 + } + return $result; + } + +} + +// 单调对列构建 +class MyQueue{ + private $queue; + + public function __construct(){ + $this->queue = new SplQueue(); //底层是双向链表实现。 + } + + public function pop($v){ + // 判断当前对列是否为空 + // 比较当前要弹出的数值是否等于队列出口元素的数值，如果相等则弹出。 + // bottom 从链表前端查看元素, dequeue 从双向链表的开头移动一个节点 + if(!$this->queue->isEmpty() && $v == $this->queue->bottom()){ + $this->queue->dequeue(); //弹出队列 + } + } + + public function push($v){ + // 判断当前对列是否为空 + // 如果push的数值大于入口元素的数值，那么就将队列后端的数值弹出，直到push的数值小于等于队列入口元素的数值为止。 + // 这样就保持了队列里的数值是单调从大到小的了。 + while (!$this->queue->isEmpty() && $v > $this->queue->top()) { + $this->queue->pop(); // pop从链表末尾弹出一个元素， + } + $this->queue->enqueue($v); + } + + // 查询当前队列里的最大值 直接返回队首 + public function max(){ + // bottom 从链表前端查看元素, top从链表末尾查看元素 + return $this->queue->bottom(); + } + + // 辅助理解: 打印队列元素 + public function println(){ + // "迭代器移动到链表头部": 可理解为从头遍历链表元素做准备。 + // 【PHP中没有指针概念，所以就没说指针。从数据结构上理解，就是把指针指向链表头部】 + $this->queue->rewind(); + + echo "Println: "; + while($this->queue->valid()){ + echo $this->queue->current()," -> "; + $this->queue->next(); + } + echo "\n"; + } +} +``` + -----------------------
diff --git a/problems/0283.移动零.md b/problems/0283.移动零.md index 9600edd3..fe8e41c1 100644 --- a/problems/0283.移动零.md +++ b/problems/0283.移动零.md @@ -133,6 +133,27 @@ var moveZeroes = function(nums) { }; ``` +TypeScript： + +```typescript +function moveZeroes(nums: number[]): void { + const length: number = nums.length; + let slowIndex: number = 0, + fastIndex: number = 0; + while (fastIndex < length) { + if (nums[fastIndex] !== 0) { + nums[slowIndex++] = nums[fastIndex]; + }; + fastIndex++; + } + while (slowIndex < length) { + nums[slowIndex++] = 0; + } +}; +``` + + + -----------------------
diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index aba6e2f3..031bc0ad 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; }; ``` @@ -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/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(); 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 c1d3515b..61eac6cd 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -137,13 +137,8 @@ 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(); } } ``` diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index e8db980d..a41a0f0a 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 @@ -298,9 +323,33 @@ 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** + ```c int wiggleMaxLength(int* nums, int numsSize){ if(numsSize <= 1) @@ -326,6 +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 @@ -375,7 +462,31 @@ 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 + } +} +``` -----------------------
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 aca9084f..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 @@ -582,7 +593,35 @@ function deleteNode(root: TreeNode | null, key: number): TreeNode | null { }; ``` +## Scala +```scala +object Solution { + def deleteNode(root: TreeNode, key: Int): TreeNode = { + if (root == null) return root // 第一种情况，没找到删除的节点，遍历到空节点直接返回 + if (root.value == key) { + // 第二种情况: 左右孩子都为空，直接删除节点，返回null + if (root.left == null && root.right == null) return null + // 第三种情况: 左孩子为空，右孩子不为空，右孩子补位 + else if (root.left == null && root.right != null) return root.right + // 第四种情况: 左孩子不为空，右孩子为空，左孩子补位 + else if (root.left != null && root.right == null) return root.left + // 第五种情况: 左右孩子都不为空，将删除节点的左子树头节点(左孩子)放到 + // 右子树的最左边节点的左孩子上，返回删除节点的右孩子 + else { + var tmp = root.right + while (tmp.left != null) tmp = tmp.left + tmp.left = root.left + return root.right + } + } + if (root.value > key) root.left = deleteNode(root.left, key) + if (root.value < key) root.right = deleteNode(root.right, key) + + root // 返回根节点，return关键字可以省略 + } +} +``` -----------------------
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 e593b876..5b75e64e 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; diff --git a/problems/0455.分发饼干.md b/problems/0455.分发饼干.md index b787dee6..f9e8e7f1 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 + } +} +``` + -----------------------
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; + } +} +``` + + -----------------------
diff --git a/problems/0491.递增子序列.md b/problems/0491.递增子序列.md index 291a19bd..080984f2 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 + } +} +``` + -----------------------
diff --git a/problems/0501.二叉搜索树中的众数.md b/problems/0501.二叉搜索树中的众数.md index f22f1eb3..671d8061 100644 --- a/problems/0501.二叉搜索树中的众数.md +++ b/problems/0501.二叉搜索树中的众数.md @@ -9,7 +9,9 @@ # 501.二叉搜索树中的众数 -[力扣题目链接](https://leetcode.cn/problems/find-mode-in-binary-search-tree/solution/) + +[力扣题目链接](https://leetcode.cn/problems/find-mode-in-binary-search-tree/) + 给定一个有相同值的二叉搜索树（BST），找出 BST 中的所有众数（出现频率最高的元素）。 @@ -798,7 +800,76 @@ function findMode(root: TreeNode | null): number[] { }; ``` +## Scala +暴力: +```scala +object Solution { + // 导包 + import scala.collection.mutable // 集合包 + import scala.util.control.Breaks.{break, breakable} // 流程控制包 + def findMode(root: TreeNode): Array[Int] = { + var map = mutable.HashMap[Int, Int]() // 存储节点的值，和该值出现的次数 + def searchBST(curNode: TreeNode): Unit = { + if (curNode == null) return + var value = map.getOrElse(curNode.value, 0) + map.put(curNode.value, value + 1) + searchBST(curNode.left) + searchBST(curNode.right) + } + searchBST(root) // 前序遍历把每个节点的值加入到里面 + // 将map转换为list，随后根据元组的第二个值进行排序 + val list = map.toList.sortWith((map1, map2) => { + if (map1._2 > map2._2) true else false + }) + var res = mutable.ArrayBuffer[Int]() + res.append(list(0)._1) // 将第一个加入结果集 + breakable { + for (i <- 1 until list.size) { + // 如果值相同就加入结果集合，反之break + if (list(i)._2 == list(0)._2) res.append(list(i)._1) + else break + } + } + res.toArray // 最终返回res的Array格式，return关键字可以省略 + } +} +``` + +递归(利用二叉搜索树的性质): +```scala +object Solution { + import scala.collection.mutable + def findMode(root: TreeNode): Array[Int] = { + var maxCount = 0 // 最大频率 + var count = 0 // 统计频率 + var pre: TreeNode = null + var result = mutable.ArrayBuffer[Int]() + + def searchBST(cur: TreeNode): Unit = { + if (cur == null) return + searchBST(cur.left) + if (pre == null) count = 1 // 等于空置为1 + else if (pre.value == cur.value) count += 1 // 与上一个节点的值相同加1 + else count = 1 // 与上一个节点的值不同 + pre = cur + + // 如果和最大值相同，则放入结果集 + if (count == maxCount) result.append(cur.value) + + // 如果当前计数大于最大值频率，更新最大值，清空结果集 + if (count > maxCount) { + maxCount = count + result.clear() + result.append(cur.value) + } + searchBST(cur.right) + } + searchBST(root) + result.toArray // return关键字可以省略 + } +} +``` ----------------------- diff --git a/problems/0509.斐波那契数.md b/problems/0509.斐波那契数.md index 71647a0a..785d0125 100644 --- a/problems/0509.斐波那契数.md +++ b/problems/0509.斐波那契数.md @@ -234,6 +234,7 @@ func fib(n int) int { } ``` ### Javascript +解法一 ```Javascript var fib = function(n) { let dp = [0, 1] @@ -244,6 +245,23 @@ var fib = function(n) { return dp[n] }; ``` +解法二：时间复杂度O(N)，空间复杂度O(1) +```Javascript +var fib = function(n) { + // 动规状态转移中，当前结果只依赖前两个元素的结果，所以只要两个变量代替dp数组记录状态过程。将空间复杂度降到O(1) + let pre1 = 1 + let pre2 = 0 + let temp + if (n === 0) return 0 + if (n === 1) return 1 + for(let i = 2; i <= n; i++) { + temp = pre1 + pre1 = pre1 + pre2 + pre2 = temp + } + return pre1 +}; +``` TypeScript diff --git a/problems/0530.二叉搜索树的最小绝对差.md b/problems/0530.二叉搜索树的最小绝对差.md index cbf8138c..809f500b 100644 --- a/problems/0530.二叉搜索树的最小绝对差.md +++ b/problems/0530.二叉搜索树的最小绝对差.md @@ -431,7 +431,84 @@ function getMinimumDifference(root: TreeNode | null): number { }; ``` +## Scala +构建二叉树的有序数组: + +```scala +object Solution { + import scala.collection.mutable + def getMinimumDifference(root: TreeNode): Int = { + val arr = mutable.ArrayBuffer[Int]() + def traversal(node: TreeNode): Unit = { + if (node == null) return + traversal(node.left) + arr.append(node.value) + traversal(node.right) + } + traversal(root) + // 在有序数组上求最小差值 + var result = Int.MaxValue + for (i <- 1 until arr.size) { + result = math.min(result, arr(i) - arr(i - 1)) + } + result // 返回最小差值 + } +} +``` + +递归记录前一个节点: + +```scala +object Solution { + def getMinimumDifference(root: TreeNode): Int = { + var result = Int.MaxValue // 初始化为最大值 + var pre: TreeNode = null // 记录前一个节点 + + def traversal(cur: TreeNode): Unit = { + if (cur == null) return + traversal(cur.left) + if (pre != null) { + // 对比result与节点之间的差值 + result = math.min(result, cur.value - pre.value) + } + pre = cur + traversal(cur.right) + } + + traversal(root) + result // return关键字可以省略 + } +} +``` + +迭代解决: + +```scala +object Solution { + import scala.collection.mutable + def getMinimumDifference(root: TreeNode): Int = { + var result = Int.MaxValue // 初始化为最大值 + var pre: TreeNode = null // 记录前一个节点 + var cur = root + var stack = mutable.Stack[TreeNode]() + while (cur != null || !stack.isEmpty) { + if (cur != null) { + stack.push(cur) + cur = cur.left + } else { + cur = stack.pop() + if (pre != null) { + result = math.min(result, cur.value - pre.value) + } + pre = cur + cur = cur.right + } + } + result // return关键字可以省略 + } +} +``` -----------------------
diff --git a/problems/0538.把二叉搜索树转换为累加树.md b/problems/0538.把二叉搜索树转换为累加树.md index 853cca6f..5c1e9e8c 100644 --- a/problems/0538.把二叉搜索树转换为累加树.md +++ b/problems/0538.把二叉搜索树转换为累加树.md @@ -352,6 +352,24 @@ function convertBST(root: TreeNode | null): TreeNode | null { }; ``` +## Scala + +```scala +object Solution { + def convertBST(root: TreeNode): TreeNode = { + var sum = 0 + def convert(node: TreeNode): Unit = { + if (node == null) return + convert(node.right) + sum += node.value + node.value = sum + convert(node.left) + } + convert(root) + root + } +} +``` ----------------------- diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md index 2a4bd3b3..7ef6463e 100644 --- a/problems/0541.反转字符串II.md +++ b/problems/0541.反转字符串II.md @@ -53,10 +53,10 @@ public: // 2. 剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符 if (i + k <= s.size()) { reverse(s.begin() + i, s.begin() + i + k ); - continue; + } else { + // 3. 剩余字符少于 k 个，则将剩余字符全部反转。 + reverse(s.begin() + i, s.end()); } - // 3. 剩余字符少于 k 个，则将剩余字符全部反转。 - reverse(s.begin() + i, s.begin() + s.size()); } return s; } @@ -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/0617.合并二叉树.md b/problems/0617.合并二叉树.md index 40aa98c3..acdcc0aa 100644 --- a/problems/0617.合并二叉树.md +++ b/problems/0617.合并二叉树.md @@ -631,7 +631,60 @@ function mergeTrees(root1: TreeNode | null, root2: TreeNode | null): TreeNode | }; ``` +## Scala +递归: +```scala +object Solution { + def mergeTrees(root1: TreeNode, root2: TreeNode): TreeNode = { + if (root1 == null) return root2 // 如果root1为空，返回root2 + if (root2 == null) return root1 // 如果root2为空，返回root1 + // 新建一个节点，值为两个节点的和 + var node = new TreeNode(root1.value + root2.value) + // 往下递归 + node.left = mergeTrees(root1.left, root2.left) + node.right = mergeTrees(root1.right, root2.right) + node // 返回node，return关键字可以省略 + } +} +``` + +迭代: +```scala +object Solution { + import scala.collection.mutable + def mergeTrees(root1: TreeNode, root2: TreeNode): TreeNode = { + if (root1 == null) return root2 + if (root2 == null) return root1 + var stack = mutable.Stack[TreeNode]() + // 先放node2再放node1 + stack.push(root2) + stack.push(root1) + while (!stack.isEmpty) { + var node1 = stack.pop() + var node2 = stack.pop() + node1.value += node2.value + if (node1.right != null && node2.right != null) { + stack.push(node2.right) + stack.push(node1.right) + } else { + if(node1.right == null){ + node1.right = node2.right + } + } + if (node1.left != null && node2.left != null) { + stack.push(node2.left) + stack.push(node1.left) + } else { + if(node1.left == null){ + node1.left = node2.left + } + } + } + root1 + } +} +``` ----------------------- diff --git a/problems/0654.最大二叉树.md b/problems/0654.最大二叉树.md index 38d2a9ec..e2a64bcd 100644 --- a/problems/0654.最大二叉树.md +++ b/problems/0654.最大二叉树.md @@ -476,7 +476,33 @@ func traversal(_ nums: inout [Int], _ left: Int, _ right: Int) -> TreeNode? { } ``` +## Scala +```scala +object Solution { + def constructMaximumBinaryTree(nums: Array[Int]): TreeNode = { + if (nums.size == 0) return null + // 找到数组最大值 + var maxIndex = 0 + var maxValue = Int.MinValue + for (i <- nums.indices) { + if (nums(i) > maxValue) { + maxIndex = i + maxValue = nums(i) + } + } + + // 构建一棵树 + var root = new TreeNode(maxValue, null, null) + + // 递归寻找左右子树 + root.left = constructMaximumBinaryTree(nums.slice(0, maxIndex)) + root.right = constructMaximumBinaryTree(nums.slice(maxIndex + 1, nums.length)) + + root // 返回root + } +} +``` -----------------------
diff --git a/problems/0669.修剪二叉搜索树.md b/problems/0669.修剪二叉搜索树.md index 154ba5a9..a286315d 100644 --- a/problems/0669.修剪二叉搜索树.md +++ b/problems/0669.修剪二叉搜索树.md @@ -453,7 +453,21 @@ function trimBST(root: TreeNode | null, low: number, high: number): TreeNode | n }; ``` +## Scala +递归法: +```scala +object Solution { + def trimBST(root: TreeNode, low: Int, high: Int): TreeNode = { + if (root == null) return null + if (root.value < low) return trimBST(root.right, low, high) + if (root.value > high) return trimBST(root.left, low, high) + root.left = trimBST(root.left, low, high) + root.right = trimBST(root.right, low, high) + root + } +} +``` ----------------------- diff --git a/problems/0700.二叉搜索树中的搜索.md b/problems/0700.二叉搜索树中的搜索.md index bda400c2..a8dcc69f 100644 --- a/problems/0700.二叉搜索树中的搜索.md +++ b/problems/0700.二叉搜索树中的搜索.md @@ -363,7 +363,34 @@ function searchBST(root: TreeNode | null, val: number): TreeNode | null { }; ``` +## Scala +递归: +```scala +object Solution { + def searchBST(root: TreeNode, value: Int): TreeNode = { + if (root == null || value == root.value) return root + // 相当于三元表达式，在Scala中if...else有返回值 + if (value < root.value) searchBST(root.left, value) else searchBST(root.right, value) + } +} +``` + +迭代: +```scala +object Solution { + def searchBST(root: TreeNode, value: Int): TreeNode = { + // 因为root是不可变量，所以需要赋值给一个可变量 + var node = root + while (node != null) { + if (value < node.value) node = node.left + else if (value > node.value) node = node.right + else return node + } + null // 没有返回就返回空 + } +} +``` -----------------------
diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md index 102f091e..06e1c88f 100644 --- a/problems/0701.二叉搜索树中的插入操作.md +++ b/problems/0701.二叉搜索树中的插入操作.md @@ -585,6 +585,43 @@ function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null { ``` +## Scala + +递归: + +```scala +object Solution { + def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = { + if (root == null) return new TreeNode(`val`) + if (`val` < root.value) root.left = insertIntoBST(root.left, `val`) + else root.right = insertIntoBST(root.right, `val`) + root // 返回根节点 + } +} +``` + +迭代: + +```scala +object Solution { + def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = { + if (root == null) { + return new TreeNode(`val`) + } + var parent = root // 记录当前节点的父节点 + var curNode = root + while (curNode != null) { + parent = curNode + if(`val` < curNode.value) curNode = curNode.left + else curNode = curNode.right + } + if(`val` < parent.value) parent.left = new TreeNode(`val`) + else parent.right = new TreeNode(`val`) + root // 最终返回根节点 + } +} +``` + -----------------------
diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index 432305e2..98b55505 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -613,6 +613,36 @@ public class Solution{ } ``` +**Kotlin:** +```kotlin +class Solution { + fun search(nums: IntArray, target: Int): Int { + // leftBorder + var left:Int = 0 + // rightBorder + var right:Int = nums.size - 1 + // 使用左闭右闭区间 + while (left <= right) { + var middle:Int = left + (right - left)/2 + // taget 在左边 + if (nums[middle] > target) { + right = middle - 1 + } + else { + // target 在右边 + if (nums[middle] < target) { + left = middle + 1 + } + // 找到了，返回 + else return middle + } + } + // 没找到，返回 + return -1 + } +} +``` + **Kotlin:** diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index b0385ce1..00886039 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -106,8 +106,9 @@ public: // 在第index个节点之前插入一个新节点，例如index为0，那么新插入的节点为链表的新头节点。 // 如果index 等于链表的长度，则说明是新插入的节点为链表的尾结点 // 如果index大于链表的长度，则返回空 + // 如果index小于0，则置为0，作为链表的新头节点。 void addAtIndex(int index, int val) { - if (index > _size) { + if (index > _size || index < 0) { return; } LinkedNode* newNode = new LinkedNode(val); diff --git a/problems/0724.寻找数组的中心索引.md b/problems/0724.寻找数组的中心索引.md index 2fc42009..5a24a884 100644 --- a/problems/0724.寻找数组的中心索引.md +++ b/problems/0724.寻找数组的中心索引.md @@ -140,6 +140,24 @@ var pivotIndex = function(nums) { }; ``` +### TypeScript + +```typescript +function pivotIndex(nums: number[]): number { + const length: number = nums.length; + const sum: number = nums.reduce((a, b) => a + b); + let leftSum: number = 0; + for (let i = 0; i < length; i++) { + const rightSum: number = sum - leftSum - nums[i]; + if (leftSum === rightSum) return i; + leftSum += nums[i]; + } + return -1; +}; +``` + + + -----------------------
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 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 c0773653..f7823e59 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; +} +``` + + + -----------------------
diff --git a/problems/0922.按奇偶排序数组II.md b/problems/0922.按奇偶排序数组II.md index 8ca46db9..49547a15 100644 --- a/problems/0922.按奇偶排序数组II.md +++ b/problems/0922.按奇偶排序数组II.md @@ -260,6 +260,75 @@ var sortArrayByParityII = function(nums) { }; ``` +### TypeScript + +> 方法一： + +```typescript +function sortArrayByParityII(nums: number[]): number[] { + const evenArr: number[] = [], + oddArr: number[] = []; + for (let num of nums) { + if (num % 2 === 0) { + evenArr.push(num); + } else { + oddArr.push(num); + } + } + const resArr: number[] = []; + for (let i = 0, length = nums.length / 2; i < length; i++) { + resArr.push(evenArr[i]); + resArr.push(oddArr[i]); + } + return resArr; +}; +``` + +> 方法二： + +```typescript +function sortArrayByParityII(nums: number[]): number[] { + const length: number = nums.length; + const resArr: number[] = []; + let evenIndex: number = 0, + oddIndex: number = 1; + for (let i = 0; i < length; i++) { + if (nums[i] % 2 === 0) { + resArr[evenIndex] = nums[i]; + evenIndex += 2; + } else { + resArr[oddIndex] = nums[i]; + oddIndex += 2; + } + } + return resArr; +}; +``` + +> 方法三： + +```typescript +function sortArrayByParityII(nums: number[]): number[] { + const length: number = nums.length; + let oddIndex: number = 1; + for (let evenIndex = 0; evenIndex < length; evenIndex += 2) { + if (nums[evenIndex] % 2 === 1) { + // 在偶数位遇到了奇数 + while (oddIndex < length && nums[oddIndex] % 2 === 1) { + oddIndex += 2; + } + // 在奇数位遇到了偶数，交换 + let temp = nums[evenIndex]; + nums[evenIndex] = nums[oddIndex]; + nums[oddIndex] = temp; + } + } + return nums; +}; +``` + + + -----------------------
diff --git a/problems/0925.长按键入.md b/problems/0925.长按键入.md index bb6aeff2..58fe99b7 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; +}; +``` + + + -----------------------
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 4052c570..a9ee232a 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 { @@ -362,6 +363,8 @@ class Solution { ``` Kotlin: + +双指针法 ```kotlin class Solution { // 双指针法 @@ -383,6 +386,32 @@ 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: @@ -420,6 +449,24 @@ object Solution { } ``` - +C#： +```csharp +public class Solution { + public int[] SortedSquares(int[] nums) { + int k = nums.Length - 1; + int[] result = new int[nums.Length]; + for (int i = 0, j = nums.Length - 1;i <= j;){ + if (nums[i] * nums[i] < nums[j] * nums[j]) { + result[k--] = nums[j] * nums[j]; + j--; + } else { + result[k--] = nums[i] * nums[i]; + i++; + } + } + return result; + } +} +``` -----------------------
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 7d160172..9691c75b 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 @@ -374,6 +374,34 @@ func removeDuplicates(_ s: String) -> String { return String(stack) } ``` + + +PHP: +```php +class Solution { + function removeDuplicates($s) { + $stack = new SplStack(); + for($i=0;$iisEmpty() || $s[$i] != $stack->top()){ + $stack->push($s[$i]); + }else{ + $stack->pop(); + } + } + + $result = ""; + while(!$stack->isEmpty()){ + $result.= $stack->top(); + $stack->pop(); + } + + // 此时字符串需要反转一下 + return strrev($result); + } +} +``` + + Scala: ```scala object Solution { @@ -396,5 +424,6 @@ object Solution { } } ``` + -----------------------
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 f4766069..46777642 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 @@ -277,6 +277,37 @@ 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]; +} +``` + TypeScript版本 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 ad9825b8..4f4cb32a 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： 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 d944ac30..7d983fbb 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) { @@ -148,6 +188,30 @@ 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; +} +``` + ----------------------- 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::() + } +} +``` + -----------------------
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::() + } +} +``` 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..e24824b9 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 @@ -356,9 +356,13 @@ func test_2_wei_bag_problem1(weight, value []int, bagweight int) int { // 递推公式 for i := 1; i < len(weight); i++ { //正序,也可以倒序 - for j := weight[i];j<= bagweight ; j++ { - dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i]) - } + for j := 0; j <= bagweight; j++ { + if j < weight[i] { + dp[i][j] = dp[i-1][j] + } else { + dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i]) + } + } } return dp[len(weight)-1][bagweight] } 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 c6d33fe1..6ea9a037 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 +双指针 + +```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) {