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317 lines
8.7 KiB
Markdown
317 lines
8.7 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20210924105952.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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## 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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return dummyHead->next;
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}
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};
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```
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## 其他语言版本
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java:
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```java
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class Solution {
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public ListNode removeNthFromEnd(ListNode head, int n) {
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ListNode dummy = new ListNode(-1);
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dummy.next = head;
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ListNode slow = dummy;
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ListNode fast = dummy;
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while (n-- > 0) {
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fast = fast.next;
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}
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// 记住 待删除节点slow 的上一节点
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ListNode prev = null;
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while (fast != null) {
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prev = slow;
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slow = slow.next;
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fast = fast.next;
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}
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// 上一节点的next指针绕过 待删除节点slow 直接指向slow的下一节点
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prev.next = slow.next;
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// 释放 待删除节点slow 的next指针, 这句删掉也能AC
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slow.next = null;
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return dummy.next;
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}
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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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head_dummy = ListNode()
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head_dummy.next = head
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slow, fast = head_dummy, head_dummy
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while(n!=0): #fast先往前走n步
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fast = fast.next
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n -= 1
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while(fast.next!=None):
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slow = slow.next
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fast = fast.next
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#fast 走到结尾后，slow的下一个节点为倒数第N个节点
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slow.next = slow.next.next #删除
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return head_dummy.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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dummyHead := &ListNode{}
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dummyHead.Next = head
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cur := head
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prev := dummyHead
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i := 1
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for cur != nil {
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cur = cur.Next
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if i > n {
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prev = prev.Next
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}
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i++
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}
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prev.Next = prev.Next.Next
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return dummyHead.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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let slowNode: ListNode | null = newHead,
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fastNode: ListNode | null = newHead;
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for (let i = 0; i < n; i++) {
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fastNode = fastNode.next;
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}
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while (fastNode.next) {
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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 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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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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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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