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.

/**
* 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;
}
};