Reverse Linked List II

Link:http://oj.leetcode.com/problems/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) {
      if (head == null || head.next == null)
            return head;
        ListNode result = new ListNode(0);
        result.next = head;
        head = result;
        ListNode pre = head, cur = head, post = head;
        int index = 1;

        while (index < m) {
            pre = pre.next;
            index++;
        }
        cur = pre.next;
        post = cur.next;
        //in this moment, index = m.
        //example, for{1,2,3,4}, and m = 1, n = 4,
        //pre is the "virtual" list node point to 1, and cur point to 1, post points to 2
        //during the iteration:
        //2->1->3->4
        //3->2->1->4
        //4->3->2->1
        while (index < n) {
            index++;
            //get the next element of post
            ListNode temp = post.next;
            //post.next should point to the next element of pre
            post.next = pre.next;
            //pre.next should point to the post now
            pre.next = post;
            //cur.next will points to the temp, and finally, cur will point to null
            cur.next = temp;
            //move post to temp
            post = temp;
        }
        return head.next;
    }
}