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) {
        ListNode result =new ListNode(0);
        result.next=head;
         ListNode cur =head;
        ListNode pre = result;
        ListNode next;
        System.out.println(result);
        int length=1;
         if(head == null || head.next == null || k < 2) return head;
         while(head.next!=null){
            length=length+1;
            head =head.next;
        }
         System.out.println(length);
        int count = length/k;
        for(int i =1;i<=count;i++){
          for(int j=1;j<k;j++){
             next = cur.next;
            cur.next=next.next;
            next.next=pre.next;
            pre.next=next;
          }
          pre = cur;
          cur = cur.next;
        }
        return result.next;
        // write code here
    }
}