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 static ListNode reverseKGroup(ListNode head, int k) {
        if (k == 1 || head == null) {
            return head;
        }
        //当前对象
        ListNode currentNode = head;
        //序号
        int i = 1;
        //返回对象
        ListNode resultNode = null;
        //上一轮的最前面那个，翻转后会在最后
        ListNode preNode = head;
        //每组的第一个节点
        ListNode start = head;
        //临时节点，用于断开连接
        ListNode temp;
        boolean isDivisible = false;
        while (currentNode != null) {
            if (i % k == 1) {
                start = currentNode;
                currentNode = currentNode.next;
                isDivisible = false;
            } else if (i % k == 0) {
                temp = currentNode.next;
                currentNode.next = null;
                currentNode = temp;
                ListNode reverseNode = reverseList(start);
                if (resultNode == null) {
                    resultNode = reverseNode;
                } else {
                    preNode.next = reverseNode;
                }
                preNode = start;
                isDivisible = true;
            } else {
                currentNode = currentNode.next;
            }
            i++;
        }
        if (!isDivisible && resultNode != null) {
            preNode.next = start;
        }
        return resultNode == null ? start : resultNode;
    }

    public static ListNode reverseList(ListNode head) {
        if (head != null && head.next != null && head.next.next != null) {
            ListNode preNode = head;
            ListNode current = head.next;
            ListNode nextNode;
            //最开始的头节点指向空
            preNode.next = null;
            while (current != null) {
                //先获取下一个节点
                nextNode = current.next;
                //当前节点指向上一个节点
                current.next = preNode;
                //上一个节点赋值为当前节点
                preNode = current;
                //当前节点往后走一步
                current = nextNode;
            }
            return preNode;
        } else if (head != null && head.next != null && head.next.next == null) {
            ListNode nextNode = head.next;
            head.next = null;
            nextNode.next = head;
            return nextNode;
        } else {
            return head;
        }
    }
}