题目描述

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：[]
示例 3：

输入：head = [1,2], n = 1
输出：[1]

使用栈来查找倒数第n个节点

/**
 * 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 dummy = new ListNode(0,head);
        Deque<ListNode> stack = new ListedLink<ListNode>();
        ListNode cur = dummy;
        while(cur!=null){
            stack.push(cur);
            cur = cur.next;
        }
        for(int i = 0;i>n;i++){
            stack.pop();
        }
        ListNode pre = stack.peek();
        pre.next= pre.next.next;
        return dummy.next;
    }
}

使用快慢指针

快慢指针相差n+1；要使得cur为null时，应该pre指针应该指向第n个节点的前一个结点。

/**
 * 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 dummy = new ListNode(0,head);
        ListNode pre = dummy;
        ListNode cur = dummy;
        for (int i = 0; i < n + 1; i++) {
            cur = cur.next;
        }
        while (cur != null) {
            pre = pre.next;
            cur = cur.next;
        }
        pre.next = pre.next.next;
        return dummy.next;
    }
}