LeetCode 141. Linked List Cycle

转载

mb5fdcae83766b7 2015-08-29 22:12:00

文章标签 LeetCode Linked List Two Pointers 快慢指针 java 文章分类 C/C++ 后端开发

原题链接在这里：https://leetcode.com/problems/linked-list-cycle/

题目：

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

题解：

快慢指针，一快runner, 每次走两步. 一慢walker, 每次走一步，若是它俩相遇就是有cycle.

Time Complexity: O(n). Space: O(1).

AC Java:

 1 /**
 2  * Definition for singly-linked list.
 3  * class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public boolean hasCycle(ListNode head) {
14         if(head == null || head.next == null){
15             return false;
16         }
17         ListNode walker = head;
18         ListNode runner = head;
19         while(runner.next != null && runner.next.next != null){
20             walker = walker.next;
21             runner = runner.next.next;
22             if(walker == runner){
23                 return true;
24             }
25         }
26         return false;
27     }
28 }

AC JavaScript:

 1 /**
 2  * Definition for singly-linked list.
 3  * function ListNode(val) {
 4  *     this.val = val;
 5  *     this.next = null;
 6  * }
 7  */
 8 
 9 /**
10  * @param {ListNode} head
11  * @return {boolean}
12  */
13 var hasCycle = function(head) {
14     if(!head || !head.next){
15         return false;
16     }
17     
18     var walker = head;
19     var runner = head;
20     while(runner && runner.next){
21         walker = walker.next;
22         runner = runner.next.next;
23         if(walker === runner){
24             return true;
25         }
26     }
27     
28     return false;
29 };