删除链表倒数第n个节点

leetcode : 19 题。

思路：定义快慢指针，让快指针先走n步，如何同时移动快慢指针，当快指针走到尾时，慢指针刚好是倒数第n个元素（的前一个）。

例：删除倒数第二个节点。n=2；

    slow   fast

      ↓     ↓ 

a->b->c->d->e->null

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode RemoveNthFromEnd(ListNode head, int n) {
        var dummyHead = new ListNode();
        dummyHead.next = head;
        var slow = dummyHead;
        var fast = dummyHead;

        //先将fast移动n步
        while(n > 0 && fast.next != null)
        {
            fast = fast.next;
            n--;
        }
        //同时移动 fast slow
        while(fast.next != null)
        {
            fast=fast.next;
            slow =slow.next;
        }

        slow.next = slow.next.next;
        return dummyHead.next;

    }
}