题解 | #链表中的节点每k个一组翻转#

import java.util.*;
public class Solution {
    public ListNode reverseKGroup (ListNode head, int k) {
         //找到每次翻转的尾部 fast-template
        ListNode tail = head;
        //遍历k次到尾部
        for (int i = 0; i < k; i++) {
            //如果不足k到了链表尾，直接返回，不翻转
            if (tail == null)
                return head;
            tail = tail.next;
        }
        //翻转时需要的前序和当前节点
        ListNode pre = null;
        ListNode cur = head;
       //在到达当前段尾节点前
        while (cur != tail) {
            //翻转
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
       //当前尾指向下一段要翻转的链表
        head.next = reverseKGroup(tail, k);
        return pre;
   }
}