Remove Nth Node From End of List [LeetCode]

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

Summary: Be careful about corner cases, like n = 1,  or size of linked list equals to n.

     ListNode *removeNthFromEnd(ListNode *head, int n) {
         vector<ListNode *> cache;  // size should be n + 1 or n (in this case, size of linked list is n)
         ListNode * current = head;
         while(current != NULL){
             cache.push_back(current);
             if(cache.size() > n + )
                 cache.erase(cache.begin());
             current = current -> next;
         }

         if(cache.size() == n + ){
             cache[]->next = cache[1]->next;
             return head;
         }else {
             return cache[0]->next;
         }
     }