19. 删除链表的倒数第 N 个结点
webVueBlog opened this issue · comments
I am ne zha / Jeskson commented
19. 删除链表的倒数第 N 个结点
Description
Difficulty: 中等
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]
提示:
- 链表中结点的数目为
sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
**进阶:**你能尝试使用一趟扫描实现吗?
Solution
Language: JavaScript
/**
* 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}
* 快慢指针法
*/
var removeNthFromEnd = function(head, n) {
const dummy = new ListNode(-1, head)
let [slow, fast] = [dummy, dummy]
for (let i = 0; i <= n; i++) {
fast = fast.next
}
while (fast) {
slow = slow.next
fast = fast.next
}
slow.next = slow.next.next
return dummy.next
}