Leetcode——删除链表的倒数第N个节点

Posted 2021-08-15 Yawn,

1. 题目

2. 题解

双指针遍历即可

/**
 * 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 slow = head;
        ListNode fast = head;
        for(int i = 0; i < n; i++){     //倒数第几个是从1开始算（倒数第二，其实就是前一个），而i是从0开始计数
            fast = fast.next;
        }
        if(fast == null)        
            return head.next;   //倒数n个的前一个正好是head节点
        while(fast.next != null){
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;

        return head;
    }

}