Linked List Cycle

Linked List·Easy

Given the head of a linked list, determine whether it contains a cycle. A cycle exists if some node in the list can be reached again by continuously following the next pointer.

Input/output convention for this exercise: for portability, the list is passed as an array of values plus a pos integer indicating the index where the tail's next reconnects (pos = -1 means no cycle). The provided starter includes ListNode, arrayToList, listToArray, and a _wireCycle(head, pos) helper that wires up the cycle for you. Convert the input to a real linked list, wire the cycle, then run your detection.

The classic solution is Floyd's tortoise and hare: advance two pointers at different speeds; if they ever meet, the list has a cycle.

Examples

Input: values = [3,2,0,-4], pos = 1

Output: true

Tail node (-4) reconnects to the node at index 1 (value 2), forming a cycle.

Input: values = [1,2], pos = 0

Output: true

Tail reconnects to head — a self-referential cycle.

Input: values = [1], pos = -1

Output: false

Single node, no cycle.

Constraints

0 <= values.length <= 10^4
-10^5 <= values[i] <= 10^5
pos is -1 or a valid index in [0, values.length).

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Tests

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Passed32 ms

First attempt — retry to track metric movement here.

Your approach

Pattern:Linked List
Time:O(N)
Space:O(1)

Complete approach — pattern and complexity both named.

Coach

What you got right

You correctly matched the sorted input requirement and implemented a linear-time scan using two pointers.

Where it diverged

No divergence detected. The code correctly aligns with the stated approach.

Next attempt: focus on

Try solving related sliding window problems to build familiarity with two-pointer variants.

Signals

No signals fired — clean run, you stayed in flow.

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