LeetCode 92: 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.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        // O(helper, pre)->O(2)->O(1, last)->O(3, cur)->O(4)
        if (m == n)
            return head;
            
        ListNode helper = new ListNode(0);
        helper.next = head;
        ListNode pre = helper; // pre is the node that before mth node
        
        for (int i = 1; i < m; i++)
            pre = pre.next;
        
        // last is movable, but it always points to the same node (only last.next is changed)
        ListNode last = pre.next;
        
        // cur is the node after last
        ListNode cur = last.next;
        
        int count = 0; // Number of move
        
        while (count < n-m)
        {
            last.next = cur.next;
            cur.next = pre.next;
            pre.next = cur;
            cur = last.next;
            
            count++;
        }
        
        return helper.next;
    }
}