leetcode--Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example,

Given this linked list:

1->2->3->4->5

For k = 2, you should return:

2->1->4->3->5

For k = 3, you should return:

3->2->1->4->5

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Linked List

Have you met this question in a real interview?

class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
static int number = 0;
vector<ListNode*> v;
for(ListNode *lp = head; lp!=NULL; lp=lp->next){
v.push_back(lp);
number++;
}
ListNode *result = NULL;
if(number==0||k<=1||number<k)
return head;
else if(number>=k)
{
ListNode * tail = result;
int i = 0;
for(; i<(int)number/k; i++)
{
for(int j=k*i+(k-1); j>=k*i; j--)
{
if(j==0)
{
result = v[j];
tail = result;
}
else{
tail->next = v[j];
tail = tail->next;
}
}
}
if(number%k!=0)
tail->next = v[i*k];

return result;
}
}
};

vs上运行可以，但leetcode上提交显示 runtime error??