思路：
自己画了个图
为了防止m=1的情况出现，手动在旧链表头结点前再补上一个辅助结点。

思路有了代码写起来很简单：

#include<iostream>
using namespace std;
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 (m == n || head == NULL)
			return head;
		ListNode*firstnode = new ListNode(-1);
		firstnode->next = head;

		ListNode*prev = firstnode;
		for (int i = 0; i < m - 1; i++)
			prev = prev->next;
		ListNode*curr = prev->next;
		int diff = n - m;
		
		for (int i = 0; i < diff; i++)
		{
			ListNode*temp = curr->next;
			curr->next = curr->next->next;

			temp->next = prev->next;
			prev->next = temp;
			
		}
		head = firstnode->next;
		delete firstnode;
		firstnode = NULL;
		return head;
	}
};