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 > 0)
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;
}
};