19删除链表倒数第N个节点

删除链表中倒数第N个节点

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。

1、最容易想到的,把删除倒数第N个节点转换成删除正数第M个节点

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head == null) return null;
        ListNode header = new ListNode(-1);
        header.next = head;
        //找出链表长度,问题转换成删除链表第n个节点
        int len = 1; 
        ListNode temp = head;
        while(temp.next != null){
            len++;
            temp = temp.next;
        }
        int target = len - n + 1;
        ListNode p = header;
        for(int i = 0;i < target - 1;i++,p = p.next);
        p.next = p.next.next;
        //System.out.print(p.val);
        return header.next;

    }
}

2、快慢指针

添加一个虚拟头节点header,初始是快慢指针fast、low = header,记录快慢指针之间的距离,快指针先走n步后,慢指针开始走,当fast.next == null 时,fast.next即为倒数第n个节点,删除即可

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode header = new ListNode(-1);
        header.next = head;
        ListNode fast = header;
        ListNode slow = header;
        int i = 0;//记录快慢指针的距离
        while(fast.next != null){
            if(i >= n){
                slow = slow.next;
            }
            i++;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return header.next;

    }
}
posted @ 2023-03-28 17:27  Promefire  阅读(20)  评论(0)    收藏  举报