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.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) 
    {
        if(head == NULL || head->next == NULL || m==n)
            return head;
        ListNode* he = head;
        ListNode* pre = head;
        int i=1;
        while(i<m)
        {
            he = pre;
            pre = pre->next;
            i++;
        }
        ListNode* pre_end = pre;
        ListNode* end = pre->next;
        while(i<n)
        {
            pre_end->next = end->next;
            end->next = pre;
            pre = end;
            end = pre_end->next;
            i++;
        }
        if(m==1)
            return pre;
        he->next = pre;
        return head;
    }
};

LeetCode--Reverse Linked List II