# 原地址
https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/submissions/
# 思路
快慢指针，快指针先走n步，然后快慢一起同步。当慢指针next为空，则快指针的下一个就是要删除的元素。
# 代码
```Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode realHead = new ListNode(0, head);
ListNode fast = realHead;
ListNode slow = realHead;
int i = 0;
while(fast != null){
if(i <= n){
fast = fast.next;
i++;
continue;
}
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return realHead.next;
}
}
```
19. 删除链表的倒数第N个数（链表）
