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！
## 19.删除链表的倒数第N个节点 [力扣题目链接](https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/) 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。 进阶：你能尝试使用一趟扫描实现吗？ 示例 1：  输入：head = [1,2,3,4,5], n = 2 输出：[1,2,3,5] 示例 2： 输入：head = [1], n = 1 输出：[] 示例 3： 输入：head = [1,2], n = 1 输出：[1] ## 思路 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。 思路是这样的，但要注意一些细节。 分为如下几步： * 首先这里我推荐大家使用虚拟头结点，这样方面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) * 定义fast指针和slow指针，初始值为虚拟头结点，如图：
* 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：
* fast和slow同时移动，直到fast指向末尾，如题：
* 删除slow指向的下一个节点，如图：
此时不难写出如下C++代码：
```CPP
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* dummyHead = new ListNode(0);
dummyHead->next = head;
ListNode* slow = dummyHead;
ListNode* fast = dummyHead;
while(n-- && fast != NULL) {
fast = fast->next;
}
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
while (fast != NULL) {
fast = fast->next;
slow = slow->next;
}
slow->next = slow->next->next;
return dummyHead->next;
}
};
```
## 其他语言版本
java:
```java
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode slow = dummy;
ListNode fast = dummy;
while (n-- > 0) {
fast = fast.next;
}
// 记住 待删除节点slow 的上一节点
ListNode prev = null;
while (fast != null) {
prev = slow;
slow = slow.next;
fast = fast.next;
}
// 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
prev.next = slow.next;
// 释放 待删除节点slow 的next指针, 这句删掉也能AC
slow.next = null;
return dummy.next;
}
}
```
Python:
```python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
head_dummy = ListNode()
head_dummy.next = head
slow, fast = head_dummy, head_dummy
while(n!=0): #fast先往前走n步
fast = fast.next
n -= 1
while(fast.next!=None):
slow = slow.next
fast = fast.next
#fast 走到结尾后，slow的下一个节点为倒数第N个节点
slow.next = slow.next.next #删除
return head_dummy.next
```
Go:
```Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNthFromEnd(head *ListNode, n int) *ListNode {
dummyHead := &ListNode{}
dummyHead.Next = head
cur := head
prev := dummyHead
i := 1
for cur != nil {
cur = cur.Next
if i > n {
prev = prev.Next
}
i++
}
prev.Next = prev.Next.Next
return dummyHead.Next
}
```
JavaScript:
```js
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function(head, n) {
let ret = new ListNode(0, head),
slow = fast = ret;
while(n--) fast = fast.next;
if(!fast) return ret.next;
while (fast.next) {
fast = fast.next;
slow = slow.next
};
slow.next = slow.next.next;
return ret.next;
};
```
Kotlin:
```Kotlin
fun removeNthFromEnd(head: ListNode?, n: Int): ListNode? {
val pre = ListNode(0).apply {
this.next = head
}
var fastNode: ListNode? = pre
var slowNode: ListNode? = pre
for (i in 0..n) {
fastNode = fastNode?.next
}
while (fastNode != null) {
slowNode = slowNode?.next
fastNode = fastNode.next
}
slowNode?.next = slowNode?.next?.next
return pre.next
}
```
Swift：
```swift
func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
if head == nil {
return nil
}
if n == 0 {
return head
}
let dummyHead = ListNode(-1, head)
var fast: ListNode? = dummyHead
var slow: ListNode? = dummyHead
// fast 前移 n
for _ in 0 ..< n {
fast = fast?.next
}
while fast?.next != nil {
fast = fast?.next
slow = slow?.next
}
slow?.next = slow?.next?.next
return dummyHead.next
}
```
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