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) {
        // 总数
        int count = 0;
        // 分割次数
        int splitCount = 0;
        ListNode headTemp = head;
        while (headTemp != null) {
            headTemp = headTemp.next;
            count++;
        }
        splitCount = count / k;
        headTemp = head;
        int countTemp = 0;
        while (headTemp != null) {
            // 返回第k位之后
            if (splitCount == 0) {
                break;
            }
            headTemp = reverseBetween(headTemp, k);
            splitCount -= 1;
            countTemp++;
        }
        return head;
    }

    /**
     * 反转
     *
     * @param head
     * @return
     */
    public ListNode reverseBetween(ListNode head, int k) {
        int i = 0;
        int count = 0;
        int[] tempData = new int[k];
        ListNode headTemp = head;
        ListNode result = null;
        while (headTemp != null && count <= k - 1) {
            tempData[i] = headTemp.val;
            headTemp = headTemp.next;
            i++;
            count++;
        }
        // 重置
        count = 0;
        i -= 1;
        ListNode headTemp2 = head;
        while (headTemp2 != null && count <= k - 1) {
            if (count == k - 1) {
                result = headTemp2;
            }
            headTemp2.val = tempData[i];
            headTemp2 = headTemp2.next;
            i--;
            count++;
        }
        return result.next;
    }
}