lintcode166 链表倒数第n个节点

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。

思路：设置两个指针first，second指向head，first指针先向前走n，然后两个指针一起走，first指针走到末尾，second走到倒数第n个指针！

代码

/**
 * 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 (NULL == head || n < 1) return NULL;
        ListNode *first = head;
        ListNode *second = head;
        while (n) {
            first = first->next;
            n--;
        }
        while (first) {
            first = first->next;
            second = second->next;
        }
        return second;
    }
};