Reverse Nodes in k-Group

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode reverseKGroup(ListNode head, int k) {
14         if (head == null || head.next == null)
15             return head;
16         ListNode pre = new ListNode(0);
17         pre.next = head;
18         head = pre;
19         int index = 1;
20         ListNode cur = pre.next;
21         ListNode post = cur.next;
22         //before execution, we need to check if there‘s enough element
23         while (enoughElement(cur, k)) {
24             while (index < k) {
25                 ListNode temp = post.next;
26                 post.next = pre.next;
27                 cur.next = temp;
28                 pre.next = post;
29                 post = temp;
30                 index++;
31             }
32             //after the reverse operation,we need to initial
33             //the parameter
34             index = 1;
35             pre = cur;
36             cur = cur.next;
37             //note the cur may be null
38             if (cur != null) {
39 
40                 post = cur.next;
41             }
42         }
43         return head.next;
44 
45     }
46 
47     public boolean enoughElement(ListNode head, int k) {
48         int count = 0;
49         //note the head could be null, then we cannot use head.next
50         while (head != null) {
51             head = head.next;
52             count++;
53         }
54         if (count < k)
55             return false;
56         return true;
57 
58     }
59 }