EASY NC#41 Blind #41 Linked List
141. Linked List Cycle
📖 Problem
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 (0-indexed). Note that pos is not passed as a parameter. Return true if there is a cycle in the linked list. Otherwise, return false.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
💡 Approach
- → Use Floyd's Tortoise and Hare algorithm
- → Slow pointer moves one step, fast pointer moves two steps
- → If they meet, there's a cycle; if fast reaches null, no cycle
- → Time: O(n), Space: O(1)
🛠️ Hints & Pitfalls
Hints
- •Use Floyd's Tortoise and Hare algorithm
- •Slow pointer moves one step, fast pointer moves two steps
- •If they meet, there's a cycle; if fast reaches null, no cycle
Common Pitfalls
- •Time: O(n), Space: O(1)
🧪 Test Cases
Hidden tests on submit: 1
Test Case 1
Not run Input:
hasCycle(arrayToListWithCycle([3,2,0,-4], 1)); Expected:
true Test Case 2
Not run Input:
hasCycle(arrayToListWithCycle([1,2], 0)); Expected:
true Test Case 3
Not run Input:
hasCycle(arrayToListWithCycle([1], -1)); Expected:
false 📝 Code Editor
📤 Output