Remove Nth Node From End of List（leetcode）

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.

代码：

class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
vector<ListNode*> pointers;
ListNode* temp = head;
while (temp) {
pointers.push_back(temp);
temp = temp->next;
}
pointers.push_back(NULL);
ListNode* t;
t = pointers[pointers.size()-n-1];
if (pointers.size()-1 == n)
head = head->next;
else
pointers[pointers.size()-n-2]->next = pointers[pointers.size()-n];
delete t;
return head;
}
};