Reverse Linked List II [LeetCode]

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.

Summary: First writes down the reverse parts (m to n) , then handles nodes before the m-th node and after the n-th node.

 class Solution {
 public:
     ListNode *reverseBetween(ListNode *head, int m, int n) {
         ListNode * current = head;
         ListNode * pre_m_node = NULL;
         ListNode * new_head = NULL;
         int i = ;
         while(current != NULL) {
             if(i == m -)
                 break;
             pre_m_node = current;
             i ++;
             current = current -> next;
         }
         //reverse m to n
         ListNode * pre_n_node = current;
         int num_of_reverse_op = ;
         int num_of_nodes_to_reverse = n - m + ;
         while(num_of_reverse_op < num_of_nodes_to_reverse) {
             ListNode * pre_head = new_head;
             new_head = current;
             current = current -> next;
             num_of_reverse_op ++;
             new_head -> next = pre_head;
         }
         //connect rest node after the nth node
         pre_n_node -> next = current;

         if(pre_m_node != NULL) {
             pre_m_node -> next = new_head;
             return head;
         } else {
             return new_head;
         }
     }
 };