题目

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]

解

/*** 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 reverseKGroup(ListNode head, int k) {if (head == null || head.next == null) {return head;}ListNode dummy = new ListNode(0);dummy.next = head;int count = 0;ListNode pre = dummy;ListNode cur = head;ListNode next = null;while (cur != null) {next = cur.next;count++;if (count == k) {ListNode p1 = pre.next;ListNode p2 = cur;p2.next = null;reverse(p1);pre.next = p2;p1.next = next;pre = p1;count = 0;}cur = next;}return dummy.next;}public void reverse(ListNode head) {if (head == null || head.next == null) {return;}ListNode pre = null;ListNode cur = head;ListNode next = null;while (cur != null) {next = cur.next;cur.next = pre;pre = cur;cur = next;}}
}