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.

 /**  
  * Definition for singly-linked list.  
  * struct ListNode {  
  *   int val;  
  *   ListNode *next;  
  *   ListNode(int x) : val(x), next(NULL) {}  
  * };  
  */  
   typedef ListNode ln; 

 
   ln *reverse(ln *prev, ln *cur, ln *nxt, int n, ln **lastNode)  
   {  
     int cnt = 0;  
     ln *head = cur;  
     while(cur && cnt != n+1)  
     {  
       cnt++;  
       cur->next = prev;  
       prev = cur;  
       cur = nxt;  
       if (nxt)  
         nxt = nxt->next;  
     }  
     if(cur)  
       head->next = cur;  
     if(!cur)*lastNode = prev;  
     return prev;  
   }  

   ListNode *reverseBetween(ListNode *head, int m, int n) {  
     if(!head || !head->next || (n-m<1))return head;  
     ln *prev = NULL, *cur=head, *nxt = head->next, *newHead = NULL;  
     int cnt = 1;  
     newHead = head;  
     while(cur && cnt!=m)  
     {  
       cnt++;  
       prev = cur;  
       cur = nxt;  
       if(nxt)  
         nxt = nxt->next;  
     }  
     ln *lastNode = NULL,*revHead = NULL;  
     revHead = reverse(NULL,cur,nxt, n-m, &lastNode);  
     if(prev)  
       prev->next = revHead;  
     else if(lastNode)  
       newHead = lastNode;  
     else newHead = revHead;  
     return newHead;  
   }