链表倒数第n个节点

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;
        }
    }
};