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