# Linked List Cycle

## Description

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&#x20;
>
> **Output:** true&#x20;
>
> **Explanation:** There is a cycle in the linked list, where tail connects to the second node.

<div align="left"><img src="https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png" alt=""></div>

**Example 2:**

> **Input:** head = \[1,2], pos = 0&#x20;
>
> **Output:** true&#x20;
>
> **Explanation:** There is a cycle in the linked list, where tail connects to the first node.

<div align="left"><img src="https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png" alt=""></div>

**Example 3:**

> **Input:** head = \[1], pos = -1&#x20;
>
> **Output:** false&#x20;
>
> **Explanation:** There is no cycle in the linked list.

<div align="left"><img src="https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png" alt=""></div>

**Follow up:**

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

## **Code**

```java
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        // worker 以一格的速度移动
        ListNode worker = head;
        // runner 以两格的速度移动
        ListNode runner = head;
        while (runner != null && runner.next != null) {
            runner = runner.next.next;
            worker = worker.next;
            // 如果 runner 等于 worker，说明它们相遇，即有环
            if (runner == worker) return true;
        }
        return false;
    }
}
```
