DSA TrainerLinked List Cycle
Given the head of a linked list, determine if the list contains a cycle. A cycle exists if following the next pointers eventually leads you back to a node you've already visited.
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Problem
Given head, the head of a linked list, determine if the linked list has a cycle in it.
Input
The head of a singly linked list. In the test runner, the list is described as an array and a cycle position: the tail's next pointer connects to the node at that index (-1 means no cycle).
Output
A boolean: `true` if a cycle exists, `false` if the list terminates.
Examples
Input: head = [3,2,0,-4], pos = 1
Output: true
The tail (-4) connects back to node at index 1 (value 2), forming a cycle.
Input: head = [1,2], pos = 0
Output: true
The tail (2) connects back to the head (1).
Input: head = [1], pos = -1
Output: false
Single node with no cycle.
The brute-force approach
Use a set to record every node you visit. If you ever encounter a node already in the set, there's a cycle. If you reach null, there isn't.
seen = set()
curr = head
while curr:
if curr in seen:
return True
seen.add(curr)
curr = curr.next
return FalseThis works in O(n) time but uses O(n) extra space for the set. Floyd's cycle detection (two pointers at different speeds) detects cycles in O(n) time and O(1) space.
Spotting the pattern
This is a Linked List problem. The key question to ask yourself:
If I walk through this list forever, how would I know I'm going in circles?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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