题解 | #链表中的节点每k个一组翻转# java

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

        while (head != null) {
            ListNode tail = prev;
            // 检查是否剩余节点的数量大于等于k
            for (int i = 0; i < k; i++) {
                tail = tail.next;
                if (tail == null) {
                    return dummy.next;
                }
            }

            ListNode nex = tail.next; // 记录下一个区间的起始位置
            ListNode[] reversed = reverse(head, tail); // 反转当前区间
            head = reversed[0];
            tail = reversed[1];
            // 将反转后的区间与原链表连接起来
            prev.next = head;
            tail.next = nex;
            prev = tail;
            head = tail.next;
        }

        return dummy.next;

    }
    private ListNode[] reverse(ListNode head, ListNode tail) {
        ListNode prev = tail.next;
        ListNode p = head;
        while (prev != tail) {
            ListNode nex = p.next;
            p.next = prev;
            prev = p;
            p = nex;
        }
        return new ListNode[] {tail, head};
    }
}