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docs: restructure problem paths, change section display to groups, and improve documentation content
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---
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title: '141. Linked List Cycle'
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description: Given head, the head of a linked list, determine if the linked list has a cycle in it
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icon: dot
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topics:
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- { name: "Hash Table", slug: "hash-table" }
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- { name: "Linked List", slug: "linked-list" }
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- { name: "Two Pointers", slug: "two-pointers" }
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- { name: "Floyd's Cycle Finding Algorithm", slug: "floyds-cycle-finding-algorithm" }
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video: "gBTe7lFR3vc"
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issue: "https://github.com/prdlk/leetcode/issues/79"
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sidebar:
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label: 'Linked List Cycle'
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badge: 'Easy'
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---
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::::warning
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Can you solve it using O(1) (i.e. constant) memory?
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::::
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### Example 1:
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- Input: `head = [3,2,0,-4], pos = 1`
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- Output: `true`
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- Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).
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### Example 2:
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- Input: `head = [1,2], pos = 0`
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- Output: `true`
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- Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.
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### Example 3:
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- Input: `head = [1], pos = -1`
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- Output: `false`
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- Explanation: There is no cycle in the linked list.
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### Constraints:
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- The number of the nodes in the list is in the range [0, 10^4].
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- `-10^5 <= Node.val <= 10^5`
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- pos is -1 or a valid index in the linked-list.
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## Approach
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```mermaid
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flowchart TD
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S(["hasCycle(head)"]) --> I["fast = slow = head"]
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I --> W{"fast and fast.next?"}
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W -- no --> E(["return False — ran off the end, no cycle"])
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W -- yes --> A["fast = fast.next.next — two steps"]
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A --> B["slow = slow.next — one step"]
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B --> Q{"fast is slow?"}
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Q -- yes --> R(["return True — the gap closed, so it loops"])
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Q -- no --> W
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```
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## Solution
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```py
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def hasCycle(self, head: Optional[ListNode]) -> bool:
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fast = slow = head
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while fast and fast.next:
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fast = fast.next.next
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slow = slow.next
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if fast is slow:
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return True
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return False
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```
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