# 0141. Linked List Cycle

https://leetcode.com/problems/linked-list-cycle

## Description

Given `head`, the head of a linked list, determine if the linked list has a cycle in it.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter**.

Return `true` _if there is a cycle in the linked list_. Otherwise, return `false`.

**Example 1:**

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png)

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

**Example 2:**

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png)

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

**Example 3:**

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png)

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

**Constraints:**

* The number of the nodes in the list is in the range `[0, 104]`.
* `-105 <= Node.val <= 105`
* `pos` is `-1` or a **valid index** in the linked-list.

**Follow up:** Can you solve it using `O(1)` (i.e. constant) memory?

## ac

```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) {
        // edge cases
        if (head == null || head.next == null) return false;

        // 2 pointers
        ListNode fast = head, slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) return true;
        }

        return false;
    }
}
```
