leetcode

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) {}
 * };
 */
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) return head;
		ListNode *p = head;
		std::vector<int> vec;
		while(p != NULL)
		{
			vec.push_back(p->val);
			p = p->next;
		}
		while(m < n)
		{
			std::swap(vec[m-1],vec[n-1]);
			m++;
			n--;
		}
		ListNode *node = new ListNode(0);
		ListNode *p1 = node;
		ListNode *p2 = NULL;
		for (int i = 0; i < vec.size(); i++)
		{
			p2 = new ListNode(vec[i]);
			p1->next = p2;
			p1 = p2;
		}
		return node->next;
    }
};