lintcode166 链表倒数第n个节点

Posted 2020-10-09 狗剩的美丽家园

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

 1 /**
 2  * Definition of ListNode
 3  * class ListNode {
 4  * public:
 5  *     int val;
 6  *     ListNode *next;
 7  *     ListNode(int val) {
 8  *         this->val = val;
 9  *         this->next = NULL;
10  *     }
11  * }
12  */
13 
14 
15 class Solution {
16 public:
17     /*
18      * @param head: The first node of linked list.
19      * @param n: An integer
20      * @return: Nth to last node of a singly linked list.
21      */
22     ListNode * nthToLast(ListNode * head, int n) {
23         // write your code here
24         if (NULL == head || n < 1) return NULL;
25         ListNode *first = head;
26         ListNode *second = head;
27         while (n) {
28             first = first->next;
29             n--;
30         }
31         while (first) {
32             first = first->next;
33             second = second->next;
34         }
35         return second;
36     }
37 };