- 删除链表的倒数第 N 个结点
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。
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
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function(head, n) {
let pointFront,pointBack ;
let flag=0,reArray = new ListNode();
head.front = null;
pointFront = pointBack = head;
let count = 1;
// 使单链表变成双向链表
while(pointBack.next!= null){
// pointFront始终指向的是前一个节点 pointBack始终指向的是后一个节点
pointBack = pointFront.next;
pointBack.front = pointFront;
pointFront = pointBack;
flag = 1;
}
if(flag==0) return head.next;
pointFront = pointBack.front;
// 找到最后一个结点,现在开始往回找
while(count < n){
//小于说明还没有找到
pointBack = pointFront;
pointFront = pointFront.front;
count++;
}
if(pointBack.front == null)
return pointBack.next;
else
pointFront.next = pointBack.next;
return head;
};