25. K 个一组翻转链表
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25. K 个一组翻转链表
Description
Difficulty: 困难
给你链表的头节点 head
,每 k
个节点一组进行翻转,请你返回修改后的链表。
k
是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k
的整数倍,那么请将最后剩余的节点保持原有顺序。
你不能只是单纯的改变节点内部的值,而是需要实际进行节点交换。
示例 1:
输入:head = [1,2,3,4,5], k = 2
输出:[2,1,4,3,5]
示例 2:
输入:head = [1,2,3,4,5], k = 3
输出:[3,2,1,4,5]
提示:
- 链表中的节点数目为
n
1 <= k <= n <= 5000
0 <= Node.val <= 1000
**进阶:**你可以设计一个只用 O(1)
额外内存空间的算法解决此问题吗?
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} k
* @return {ListNode}
*/
const myReverse = (head, tail) => {
let prev = tail.next;
let p = head;
while (prev !== tail) {
const nex = p.next;
p.next = prev;
prev = p;
p = nex;
}
return [tail, head];
}
var reverseKGroup = function(head, k) {
const hair = new ListNode(0)
hair.next = head
let pre = hair
while (head) {
let tail = pre
// 查看剩余部分长度是否大于等于k
for (let i = 0; i < k; ++i) {
tail = tail.next
if (!tail) {
return hair.next
}
}
const nex = tail.next;
[head, tail] = myReverse(head, tail)
// 把子链表重新接回原链表
pre.next = head
tail.next = nex
pre = tail
head = tail.next
}
return hair.next
};