[LeetCode] 141. Linked List Cycle

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

环形链表。

题意很简单，思路是用快慢指针，慢指针每走一步，快指针走两步。如果快慢指针在某个地方相遇了，说明有环；否则快指针就会遍历到链表尾部从而会退出循环。

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {ListNode} head
 3  * @return {boolean}
 4  */
 5 var hasCycle = function(head) {
 6     // corner case
 7     if (head === null || head.next === null) {
 8         return false;
 9     }
10 
11     // normal case
12     let slow = head;
13     let fast = head;
14     while (fast !== null && fast.next !== null) {
15         slow = slow.next;
16         fast = fast.next.next;
17         if (slow === fast) {
18             return true;
19         }
20     }
21     return false;
22 };

Java实现

 1 public class Solution {
 2     public boolean hasCycle(ListNode head) {
 3         // corner case
 4         if (head == null || head.next == null) {
 5             return false;
 6         }
 7 
 8         // normal case
 9         ListNode slow = head;
10         ListNode fast = head;
11         while (fast != null && fast.next != null) {
12             slow = slow.next;
13             fast = fast.next.next;
14             if (slow == fast) {
15                 return true;
16             }
17         }
18         return false;
19     }
20 }

LeetCode 题目总结