题解 | #链表中的节点每k个一组翻转#

import java.util.*;
public class Solution {
    public ListNode reverseKGroup (ListNode head, int k) {
        if (head == null || head.next == null) return head;
        if (k <= 1) return head;
        ListNode left = new ListNode(-1);
        left.next = head;
        ListNode pre = left;
        while (pre != null) {
            pre = reverse_next_k(pre, k);
        }
        return left.next;
    }
    
    
    /**
     * 翻转前k个元素，不足k个则保持原样
     * 返回本次翻转的k个元素呆的最有一个元素，作为下一次翻转的头部，方便采用前插
     */
    public static ListNode reverse_next_k(ListNode head, int k) {
        ListNode right = head.next;
        for (int i = 1; i <= k; i++) {
            // 不足k个则直接返回头部
            if (right == null) return null;
            right = right.next;
        }
        ListNode cur = head.next;
        while (cur.next != right) {
            ListNode next = cur.next;
            cur.next = next.next;
            next.next = head.next;
            head.next = next;
        }
        return cur;
    }
}