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
        // 解题思路：
        // 1.k=1或不足k个 或者为null，则原样返回head
        // 2.遍历找到第k+1个节点
        // 3.对前k个节点反转
        // 4.递归 从第k+1个节点开始，继续进行k反转
        // 注意：原来的head 就是反转后的末尾，pre 就是开头

        if (k == 1 || head == null) {
            return head;
        }

        ListNode root = new ListNode(-1);
        ListNode nextHead = head;

        for (int i = 0; i < k; i++) {
            if (nextHead == null) {
                return head;
            }
            nextHead = nextHead.next;
        }

        ListNode pre = null;
        ListNode cur = head;
        ListNode next;

        while ( cur != nextHead) {
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }

        root.next = pre;
        head.next = reverseKGroup (nextHead,  k);

        return root.next;

    }
}