19\. 删除链表的倒数第 N 个结点
# 19. 删除链表的倒数第 N 个结点 (opens new window)
# Description
Difficulty: 中等
Related Topics: 链表 (opens new window), 双指针 (opens new window)
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]
1
2
2
示例 2:
输入:head = [1], n = 1
输出:[]
1
2
2
示例 3:
输入:head = [1,2], n = 1
输出:[1]
1
2
2
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}
* 1.暴力求解
* 找到列表的长度
* 删除从列表开头数起第(L-n+1)个节点
*
* 2.快慢指针法
* 关键字:倒数第N个
* 模式识别:
* 涉及链表的特殊位置,考虑快慢指针
* 要删除链表节点,找到它的前驱
*
* 双指针
*/
var removeNthFromEnd = function(head, n) {
const dummy = new ListNode(0, 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
}
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