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
     */

    private ListNode reverse(ListNode head, ListNode tail) {
        ListNode pre = null;
        ListNode cur = head;
        while (cur != tail){
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
            return pre;//此时pre是tail前一个节点。反转区间最后一个，也就是该组反转区间的新头节点

    }
    public ListNode reverseKGroup (ListNode head, int k) {
        ListNode cur = head;
        for (int i = 0; i < k; i++) {
            if (cur == null) return head;
            cur = cur.next;
        }
        //翻转区间的新的头节点
        ListNode newhead = reverse(head, cur);
        head.next = reverseKGroup(cur, k);

        //连接各组
        return newhead;



    }
}