[Leetcode] Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

Solution:

 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 reverseBetween(ListNode head, int m, int n) {
14         if(m==n)
15             return head;
16         ListNode dummy=new ListNode(-1);
17         dummy.next=head;
18         ListNode pointer=dummy;
19         int cnt=m-1;
20         while(cnt!=0){
21             pointer=pointer.next;
22             cnt--;
23         }
24         ListNode l1,l2,l3;
25         l1=pointer.next;
26         ListNode last=l1;
27         l2=l1.next;
28         l3=l2.next;
29         cnt=n-m-1;
30         while(cnt!=0){
31             l2.next=l1;
32             
33             l1=l2;
34             l2=l3;
35             l3=l3.next;
36             cnt--;
37         }
38         l2.next=l1;
39         last.next=l3;
40         pointer.next=l2;
41         return dummy.next;
42     }
43 }