题目

https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/

代码

A-B-C-D-E-F
len=6
倒数第n位=2
正着数就是m位=4

快指针先走n位 然后快慢一起走，快走到头的时候就走了m步了 这个时候慢指针刚好就到需要删除的位置了

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(-1);
        dummy.next=head;
        
        ListNode fast = dummy;
        ListNode slow= dummy;
        
        for(int i=1;i<=n+1;i++){
            fast=fast.next;
        }
        
        while(fast!=null){
            fast=fast.next;
            slow=slow.next;
        }
        
        slow.next=slow.next.next;
        return dummy.next;
    }
}