leetcode-Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

思路：方法与输出倒数第n个结点类似，先找到倒数第N+1个结点p,p->next=p->next->next；

代码：ListNode *removeNthFromEnd(ListNode *head, int n) {
    ListNode *p=head;
ListNode *q=head;
while(n>0)
{
q=q->next;
--n;
}
if(q == NULL)
{

          delete head;
   return head->next;
}
while(q->next!=NULL)
{
p=p->next;
q=q->next;
}

 delete p->next;
p->next=p->next->next;
return head;
    }