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;
    }
};