题解 | #牛群的重新分组#

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 dummy = new ListNode(0);
        dummy.next = head;

        ListNode prev = dummy;
        ListNode curr = head;

        int count = 0;
        while (curr != null) {
            count++;
            if (count % k == 0) {
                prev = reverse(prev, curr.next);
                curr = prev.next;
            } else {
                curr = curr.next;
            }
        }

        return dummy.next;
    }
    ListNode reverse(ListNode prev, ListNode end) {
        ListNode curr = prev.next;
        ListNode tail = curr;

        while (curr != end) {
            ListNode temp = curr.next;
            curr.next = prev.next;
            prev.next = curr;
            curr = temp;
        }

        tail.next = end;
        return tail;
    }
}

解题思路分析：构造虚拟头结点为初始prev，初始curr指向head。用count++计数，不能整除k就curr后移一位，能整除就带入函数操作，再更新prev和curr指针；写了一个反转两个节点的函数。