19. 删除链表的倒数第 N 个结点

小半 • 2023年1月4日下午2:16 • 数据结构 • 阅读 160

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

难度中等1519

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]

提示：

链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

通过次数479,807提交次数1,122,789

/**
 * 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 tail = head;
        int num =0;
        //链表计数
        while(tail.next!=null)
        {
            tail = tail.next;
            num++;
        }
        if(n == num+1) return head = head.next;
        num = num-n;
        tail = head;
        while(num>0)
        {
            num--;
            tail = tail.next;
        }
        tail.next = tail.next.next;
        return head;
    }
}

/**
 * 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 head1 = new ListNode(0,head);
        ListNode first = head1;
        ListNode seconde = head1;
        while(first.next!=null)
        {
            first = first.next;
            n--;
            if(n<0) seconde = seconde.next; 
        }
        seconde.next = seconde.next.next;
        return head1.next;
    }
}

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19. 删除链表的倒数第 N 个结点

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