题解 | #牛群的重新分组#

考察链表的相关操作。

创建一个头节点并指向头结点，随后根据链表的长度确定翻转的组数。

在每一组内进行链表翻转随后连接相邻的组。最后返回虚拟头节点的下一个节点。

完整Java代码如下所示

import java.util.*;

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

public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;

        ListNode prev = dummy;
        ListNode curr = head;

        int length = 0;
        while (head != null) {
            length++;
            head = head.next;
        }

        for (int i = 0; i < length / k; i++) {
            for (int j = 1; j < k; j++) {
                ListNode next = curr.next;
                curr.next = next.next;
                next.next = prev.next;
                prev.next = next;
            }
            prev = curr;
            curr = curr.next;
        }

        return dummy.next;
    }
}