mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2026-02-02 18:39:09 +08:00
447 lines
12 KiB
Markdown
447 lines
12 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，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！</strong></p>
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# 19.删除链表的倒数第N个节点
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[力扣题目链接](https://leetcode.cn/problems/remove-nth-node-from-end-of-list/)
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给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。
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进阶：你能尝试使用一趟扫描实现吗？
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示例 1：
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输入：head = [1,2,3,4,5], n = 2
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输出：[1,2,3,5]
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示例 2：
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输入：head = [1], n = 1
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输出：[]
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示例 3：
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输入：head = [1,2], n = 1
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输出：[1]
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## 算法公开课
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：：[链表遍历学清楚！ | LeetCode：19.删除链表倒数第N个节点](https://www.bilibili.com/video/BV1vW4y1U7Gf)，相信结合视频再看本篇题解，更有助于大家对链表的理解。**
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## 思路
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双指针的经典应用，如果要删除倒数第n个节点，让fast移动n步，然后让fast和slow同时移动，直到fast指向链表末尾。删掉slow所指向的节点就可以了。
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思路是这样的，但要注意一些细节。
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分为如下几步：
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* 首先这里我推荐大家使用虚拟头结点，这样方便处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： [链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html)
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* 定义fast指针和slow指针，初始值为虚拟头结点，如图：
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<img src='https://code-thinking.cdn.bcebos.com/pics/19.%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E7%9A%84%E5%80%92%E6%95%B0%E7%AC%ACN%E4%B8%AA%E8%8A%82%E7%82%B9.png' width=600> </img></div>
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* fast首先走n + 1步 ，为什么是n+1呢，因为只有这样同时移动的时候slow才能指向删除节点的上一个节点（方便做删除操作），如图：
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<img src='https://code-thinking.cdn.bcebos.com/pics/19.%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E7%9A%84%E5%80%92%E6%95%B0%E7%AC%ACN%E4%B8%AA%E8%8A%82%E7%82%B91.png' width=600> </img></div>
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* fast和slow同时移动，直到fast指向末尾，如题：
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<img src='https://code-thinking.cdn.bcebos.com/pics/19.%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E7%9A%84%E5%80%92%E6%95%B0%E7%AC%ACN%E4%B8%AA%E8%8A%82%E7%82%B92.png' width=600> </img></div>
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* 删除slow指向的下一个节点，如图：
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<img src='https://code-thinking.cdn.bcebos.com/pics/19.%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E7%9A%84%E5%80%92%E6%95%B0%E7%AC%ACN%E4%B8%AA%E8%8A%82%E7%82%B93.png' width=600> </img></div>
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此时不难写出如下C++代码：
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```CPP
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class Solution {
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public:
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ListNode* removeNthFromEnd(ListNode* head, int n) {
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ListNode* dummyHead = new ListNode(0);
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dummyHead->next = head;
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ListNode* slow = dummyHead;
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ListNode* fast = dummyHead;
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while(n-- && fast != NULL) {
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fast = fast->next;
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}
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fast = fast->next; // fast再提前走一步，因为需要让slow指向删除节点的上一个节点
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while (fast != NULL) {
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fast = fast->next;
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slow = slow->next;
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}
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slow->next = slow->next->next;
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// ListNode *tmp = slow->next; C++释放内存的逻辑
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// slow->next = tmp->next;
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// delete tmp;
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return dummyHead->next;
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}
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};
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```
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* 时间复杂度: O(n)
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* 空间复杂度: O(1)
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## 其他语言版本
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### Java:
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```java
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public ListNode removeNthFromEnd(ListNode head, int n){
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ListNode dummyNode = new ListNode(0);
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dummyNode.next = head;
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ListNode fastIndex = dummyNode;
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ListNode slowIndex = dummyNode;
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// 只要快慢指针相差 n 个结点即可
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for (int i = 0; i < n ; i++){
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fastIndex = fastIndex.next;
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}
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while (fastIndex != null){
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fastIndex = fastIndex.next;
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slowIndex = slowIndex.next;
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}
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//此时 slowIndex 的位置就是待删除元素的前一个位置。
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//具体情况可自己画一个链表长度为 3 的图来模拟代码来理解
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slowIndex.next = slowIndex.next.next;
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return dummyNode.next;
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}
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```
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### Python:
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
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# 创建一个虚拟节点，并将其下一个指针设置为链表的头部
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dummy_head = ListNode(0, head)
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# 创建两个指针，慢指针和快指针，并将它们初始化为虚拟节点
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slow = fast = dummy_head
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# 快指针比慢指针快 n+1 步
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for i in range(n+1):
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fast = fast.next
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# 移动两个指针，直到快速指针到达链表的末尾
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while fast:
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slow = slow.next
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fast = fast.next
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# 通过更新第 (n-1) 个节点的 next 指针删除第 n 个节点
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slow.next = slow.next.next
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return dummy_head.next
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```
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### Go:
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```Go
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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func removeNthFromEnd(head *ListNode, n int) *ListNode {
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dummyNode := &ListNode{0, head}
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fast, slow := dummyNode, dummyNode
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for i := 0; i <= n; i++ { // 注意<=，否则快指针为空时，慢指针正好在倒数第n个上面
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fast = fast.Next
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}
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for fast != nil {
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fast = fast.Next
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slow = slow.Next
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}
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slow.Next = slow.Next.Next
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return dummyNode.Next
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}
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```
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### JavaScript:
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```js
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/**
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* @param {ListNode} head
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* @param {number} n
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* @return {ListNode}
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*/
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var removeNthFromEnd = function(head, n) {
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let ret = new ListNode(0, head),
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slow = fast = ret;
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while(n--) fast = fast.next;
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while (fast.next !== null) {
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fast = fast.next;
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slow = slow.next
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};
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slow.next = slow.next.next;
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return ret.next;
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};
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```
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### TypeScript:
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版本一（快慢指针法）：
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```typescript
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function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
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let newHead: ListNode | null = new ListNode(0, head);
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//根据leetcode题目的定义可推断这里快慢指针均不需要定义为ListNode | null。
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let slowNode: ListNode = newHead;
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let fastNode: ListNode = newHead;
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while(n--) {
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fastNode = fastNode.next!; //由虚拟头节点前进n个节点时,fastNode.next可推断不为null。
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}
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while(fastNode.next) { //遍历直至fastNode.next = null， 即尾部节点。 此时slowNode指向倒数第n个节点。
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fastNode = fastNode.next;
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slowNode = slowNode.next!;
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}
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slowNode.next = slowNode.next!.next; //倒数第n个节点可推断其next节点不为空。
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return newHead.next;
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}
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```
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版本二（计算节点总数法）：
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```typescript
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function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
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let curNode: ListNode | null = head;
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let listSize: number = 0;
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while (curNode) {
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curNode = curNode.next;
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listSize++;
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}
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if (listSize === n) {
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head = head.next;
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} else {
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curNode = head;
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for (let i = 0; i < listSize - n - 1; i++) {
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curNode = curNode.next;
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}
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curNode.next = curNode.next.next;
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}
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return head;
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};
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```
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版本三（递归倒退n法）：
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```typescript
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function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
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let newHead: ListNode | null = new ListNode(0, head);
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let cnt = 0;
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function recur(node) {
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if (node === null) return;
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recur(node.next);
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cnt++;
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if (cnt === n + 1) {
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node.next = node.next.next;
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}
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}
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recur(newHead);
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return newHead.next;
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};
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```
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### Kotlin:
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```Kotlin
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fun removeNthFromEnd(head: ListNode?, n: Int): ListNode? {
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val pre = ListNode(0).apply {
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this.next = head
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}
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var fastNode: ListNode? = pre
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var slowNode: ListNode? = pre
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for (i in 0..n) {
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fastNode = fastNode?.next
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}
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while (fastNode != null) {
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slowNode = slowNode?.next
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fastNode = fastNode.next
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}
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slowNode?.next = slowNode?.next?.next
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return pre.next
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}
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```
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### Swift：
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```swift
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func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
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if head == nil {
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return nil
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}
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if n == 0 {
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return head
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}
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let dummyHead = ListNode(-1, head)
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var fast: ListNode? = dummyHead
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var slow: ListNode? = dummyHead
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// fast 前移 n
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for _ in 0 ..< n {
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fast = fast?.next
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}
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while fast?.next != nil {
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fast = fast?.next
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slow = slow?.next
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}
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slow?.next = slow?.next?.next
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return dummyHead.next
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}
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```
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### PHP:
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```php
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function removeNthFromEnd($head, $n) {
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// 设置虚拟头节点
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$dummyHead = new ListNode();
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$dummyHead->next = $head;
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$slow = $fast = $dummyHead;
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while($n-- && $fast != null){
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$fast = $fast->next;
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}
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// fast 再走一步，让 slow 指向删除节点的上一个节点
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$fast = $fast->next;
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while ($fast != NULL) {
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$fast = $fast->next;
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$slow = $slow->next;
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}
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$slow->next = $slow->next->next;
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return $dummyHead->next;
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}
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```
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### Scala:
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```scala
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object Solution {
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def removeNthFromEnd(head: ListNode, n: Int): ListNode = {
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val dummy = new ListNode(-1, head) // 定义虚拟头节点
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var fast = head // 快指针从头开始走
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var slow = dummy // 慢指针从虚拟头开始头
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// 因为参数 n 是不可变量，所以不能使用 while(n>0){n-=1}的方式
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for (i <- 0 until n) {
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fast = fast.next
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}
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// 快指针和满指针一起走，直到fast走到null
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while (fast != null) {
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slow = slow.next
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fast = fast.next
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}
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// 删除slow的下一个节点
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slow.next = slow.next.next
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// 返回虚拟头节点的下一个
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dummy.next
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}
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}
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```
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### Rust:
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```rust
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impl Solution {
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pub fn remove_nth_from_end(head: Option<Box<ListNode>>, mut n: i32) -> Option<Box<ListNode>> {
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let mut dummy_head = Box::new(ListNode::new(0));
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dummy_head.next = head;
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let mut fast = &dummy_head.clone();
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let mut slow = &mut dummy_head;
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while n > 0 {
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fast = fast.next.as_ref().unwrap();
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n -= 1;
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}
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while fast.next.is_some() {
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fast = fast.next.as_ref().unwrap();
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slow = slow.next.as_mut().unwrap();
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}
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slow.next = slow.next.as_mut().unwrap().next.take();
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dummy_head.next
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}
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}
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```
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### C：
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```c
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/**c语言单链表的定义
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* struct ListNode *next;
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* };
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*/
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struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
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//定义虚拟头节点dummy 并初始化使其指向head
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struct ListNode* dummy = malloc(sizeof(struct ListNode));
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dummy->val = 0;
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dummy->next = head;
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//定义 fast slow 双指针
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struct ListNode* fast = head;
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struct ListNode* slow = dummy;
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for (int i = 0; i < n; ++i) {
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fast = fast->next;
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}
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while (fast) {
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fast = fast->next;
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slow = slow->next;
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}
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slow->next = slow->next->next;//删除倒数第n个节点
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head = dummy->next;
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free(dummy);//删除虚拟节点dummy
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return head;
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}
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```
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### C#：
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```csharp
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public class Solution {
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public ListNode RemoveNthFromEnd(ListNode head, int n) {
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ListNode dummpHead = new ListNode(0);
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dummpHead.next = head;
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var fastNode = dummpHead;
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var slowNode = dummpHead;
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while(n-- != 0 && fastNode != null)
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{
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fastNode = fastNode.next;
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}
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while(fastNode.next != null)
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{
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fastNode = fastNode.next;
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slowNode = slowNode.next;
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}
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slowNode.next = slowNode.next.next;
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return dummpHead.next;
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}
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}
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```
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<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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||
</a>
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