[leetcode]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.

注意题目 Do it in-place and in one-pass.

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int count = 1;
        ListNode * tmp = head;
        ListNode * subListHead;
        ListNode * subListTail;
        ListNode * temp;
        ListNode * pre = NULL;
        ListNode * breakpoint1;
        ListNode * breakpoint2;
        while(count <= n)
        {
            temp = tmp->next;
            if(count == m)
            {
                subListTail = tmp;
                breakpoint1 = pre;
            }
            if(count == n)
            {
                subListHead = tmp;
                breakpoint2 = tmp->next;
            }
            if(count > m)
            {
                tmp->next = pre;
            }
            pre = tmp;
            tmp = temp;
            count++;
        }
        ListNode * newHead;
        if(breakpoint1 == NULL)
        {
            newHead = subListHead;
        }
        else
        {
            newHead = head;
            breakpoint1->next = subListHead;
        }
        subListTail->next = breakpoint2;
        return newHead;
    }
};