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：

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode* removeNthFromEnd(ListNode* head, int n) {12         ListNode * p, * q, * pPre;13         pPre = NULL;14         p = q = head;15         while(--n > 0)16             q = q->next;17         while(q->next){18             pPre = p;19             p = p->next;20             q = q->next;        }21         if(pPre == NULL){22             head = p->next;23             delete p;24         }else{25             pPre->next = p->next;26             delete p;27         }28         return head;29     }30 };