算法描述：

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

解題思路：鏈表題，首先要畫圖。四個指針，頭指針，前指針，臨時指針。

    ListNode* reverseBetween(ListNode* head, int m, int
 
 n) {
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* prev = dummy;
        ListNode* cur = dummy;
        for(int i=1; i < m; i++) 
            prev = prev->next;
        cur = prev->next;
        for(int i =m; i<n; i++){
            ListNode
 
* temp = cur->next;
            cur->next = temp->next;
            temp->next =prev->next;
            prev->next = temp;
            
         }
        return dummy->next;
    }

LeetCode-92-Reverse Linked List II