# 141. Linked List Cycle

#### Easy

***

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.

&#x20;

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

```python
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        slow = head
        fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if fast == slow:
                return True
        return False
```