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.

 1 class Solution {
 2 public:
 3     ListNode *reverseBetween(ListNode *head, int m, int n) {
 4         ListNode * current = head;
 5         ListNode * pre_m_node = NULL;
 6         ListNode * new_head = NULL;
 7         int i = 0;
 8         while(current != NULL) {
 9             if(i == m -1)
10                 break;
11             pre_m_node = current;
12             i ++;
13             current = current -> next;
14         }
15         //reverse m to n
16         ListNode * pre_n_node = current; 
17         int num_of_reverse_op = 0;
18         int num_of_nodes_to_reverse = n - m + 1;
19         while(num_of_reverse_op < num_of_nodes_to_reverse) {
20             ListNode * pre_head = new_head;
21             new_head = current;
22             current = current -> next;
23             num_of_reverse_op ++;
24             new_head -> next = pre_head;
25         }
26         //connect rest node after the nth node
27         pre_n_node -> next = current;
28         
29         if(pre_m_node != NULL) {
30             pre_m_node -> next = new_head;
31             return head;
32         } else {
33             return new_head;
34         }
35     }
36 };