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Leetcode 19. 删除链表的倒数第N个结点(Remove nth node from end of list)

发布时间 2023-08-16 23:38:57作者: Ahci

题目链接

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.

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：[]

提示:

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

思路

暴力解法: 可以先用循环遍历一次链表,获取长度. 再次循环使指针指向要删除的节点的前一个节点进行删除.
双指针法: 创建快指针(right)和慢指针(left), 首先使快指针移动n + 1步, 再让两个指针同时移动, 直到right == null, 此时慢指针指向的就是要删除的节点的前一个节点.

代码实现

暴力解法:

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        int length = getLength(head);
        ListNode dummy = new ListNode(-1, head);
        ListNode current = dummy;

        for(int i = 0; i < length - n; i++) {
            current = current.next;
        }

        current.next = current.next.next;
        return dummy.next;

    }

    public int getLength(ListNode head) {
        int length = 0;
        while(head != null) {
            head = head.next;
            length++;
        }

        return length;
    }
}

双指针法:

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(-1, head);
        ListNode left = dummy;
        ListNode right = dummy;

        for(int i = 0; i < n; i++) {
            right = right.next;
        }

        while(right.next != null) {
            right = right.next;
            left = left.next;
        }

        left.next = left.next.next;
        return dummy.next;
    }
}