题解 | #链表中的节点每k个一组翻转#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    public ListNode reverseKGroup (ListNode head, int k) {
        // write code here
        ListNode tail = head;
        for(int i = 0; i < k; i++){
            //如果剩余长度不足K,则直接返回
            if(tail == null){
                return head;
            }
            tail = tail.next;
        }    

        //头插法，pre指向新链表的头节点
        ListNode pre = null;
        ListNode cur = head;
        while(cur != tail){
            ListNode tmp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = tmp;
        }
        //当前尾指向下一段要反转的链表
        head.next = reverseKGroup(cur, k);
        return pre;
    }
}