https://leetcode.com/problems/reverse-linked-list-ii/

难度还好，和之前的反转链表没本质区别，把头尾拆出来即可

关于stack存储，和汪童鞋中午吃饭讨论说起来 存listnode肯定更好一些

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
/**
 * 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) {
        stack<ListNode*>s;
        ListNode* head0 = new ListNode(0);
        head0->next = head;
        ListNode* pLeft = head0;
        ListNode* pRight = head0;
        for(int i = 1; i < m ;i++) {//[1,2)
            pLeft = pLeft->next;//1
            pRight = pRight->next;//1
        }
        for(int i = m;i <= n ;i++){//[2,4]
            pRight = pRight->next;//2-3-4
            s.push(pRight);
           // printf("val= %d\n",pRight->val);
        }
        pRight = pRight->next;
        while(!s.empty()) {
            pLeft ->next = s.top();
            s.pop();
            pLeft = pLeft->next;
        }
        pLeft->next = pRight;
        return head0->next;
    }
};