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 = 0;
        while(current != NULL) {
            if(i == m -1)
                break;
            pre_m_node = current;
            i ++;
            current = current -> next;
        }
        //reverse m to n
        ListNode * pre_n_node = current; 
        int num_of_reverse_op = 0;
        int num_of_nodes_to_reverse = n - m + 1;
        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;
        }
    }
};