leetcode/apps/docs/content/(hash-table)/141-linked-list-cycle.mdx

---
title: '141. Linked List Cycle'
description: Given head, the head of a linked list, determine if the linked list has a cycle in it
sidebar:
  label: 'Linked List Cycle'
  badge: 'Easy'
---

<Badge variant="accent">Linked List</Badge>

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

### Example 1:
- 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:
- 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:
- 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, 10^4].
- `-10^5 <= Node.val <= 10^5`
- pos is -1 or a valid index in the linked-list.

## Approach

```mermaid
flowchart TD
  S(["hasCycle(head)"]) --> I["fast = slow = head"]
  I --> W{"fast and fast.next?"}
  W -- no --> E(["return False — ran off the end, no cycle"])
  W -- yes --> A["fast = fast.next.next — two steps"]
  A --> B["slow = slow.next — one step"]
  B --> Q{"fast is slow?"}
  Q -- yes --> R(["return True — the gap closed, so it loops"])
  Q -- no --> W
```

## Solution

```py
# 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:
        fast = slow = head

        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast is slow:
                return True
        return False
```

## Explanation

<YouTube id="gBTe7lFR3vc" title="Linked List Cycle - Floyd's Tortoise and Hare - Leetcode 141 - Python" />