找到单链表倒数第n个节点，保证链表中节点的最少数量为n。

样例

给出链表 3->2->1->5->null和n = 2，返回倒数第二个节点的值1.

/** * Definition of ListNode * class ListNode { * public: *     int val; *     ListNode *next; *     ListNode(int val) { *         this->val = val; *         this->next = NULL; *     } * } */class Solution {public:    /**     * @param head: The first node of linked list.     * @param n: An integer.     * @return: Nth to last node of a singly linked list.      */    ListNode *nthToLast(ListNode *head, int n) {        // write your code here        if(!head||!head->next) return head;        int res=0;        ListNode *p=head;        while(p){            res++;            p=p->next;        }        ListNode *q=head;        for(int i=1;i<=res;i++){            if(res-i+1==n){                return q;            }            q=q->next;        }    }};