LeetCode OJ：Remove Nth Node From End of List（倒序移除List中的元素）

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.

只能走一遍，要求移除倒着数的第n个元素，开始想用递归做，但是一直失败，没办法看了一下别人的答案，如果总数为N那么倒着第n个相当于正着第N-n个，N的计数通过一次遍历计数即相对的可以得到，代码如下：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* removeNthFromEnd(ListNode* head, int n) {
         ListNode * p, * q, * pPre;
         pPre = NULL;
         p = q = head;
         while(--n > )
             q = q->next;
         while(q->next){
             pPre = p;
             p = p->next;
             q = q->next;        }
         if(pPre == NULL){
             head = p->next;
             delete p;
         }else{
             pPre->next = p->next;
             delete p;
         }
         return head;
     }
 };