题解 | #牛群的重新分组#

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
        ListNode result = null;
        ListNode tempNode = null;
        int i = 0;
        while(head != null){
            i++;
            ListNode temp = head.next;
            head.next = tempNode;
            tempNode = head;
            head = temp;
            if(i == k){
                // 每 k 个节点一组进行翻转
               ListNode reverseGroupresult =  reverse(tempNode);
               ListNode curNode = reverseGroupresult;
               for(int j = 1;j < k;j++){
                    curNode = curNode.next;
               }
               curNode.next = result;
               result = reverseGroupresult;
                i = 0;
                tempNode = null;
            }
        }
        // 最后剩余的节点保持原有顺序
        if(tempNode != null){
            ListNode curNode = tempNode;
            while(curNode.next != null){
                curNode = curNode.next;
            }
            curNode.next = result;
            result = tempNode;
        }
        return reverse(result);
    }

    // 反转链表
    public ListNode reverse(ListNode head){
        ListNode result = null;
        while(head != null){
            ListNode tempNode = head.next;
            head.next = result;
            result = head;
            head = tempNode;
        }
        return result;
    }
}