View this problem on Leetcode.com

🧩 Problem:

Given a singly linked list, determine if it contains a cycle.

🧪 Example:

Input:
head = [3,2,0,-4], pos = 1 (cycle back to node with value 2)
Output: True

Input:
head = [1,2], pos = -1 (no cycle)
Output: False

🧠 How to Approach It

❌ Brute force would use a set() to track visited nodes — O(n) space.

✅ Instead, we use Floyd’s Cycle Detection Algorithm (Tortoise & Hare):

• 🐢 slow pointer moves 1 step

• 🐇 fast pointer moves 2 steps

• If they meet: cycle exists

• If fast or fast.next is None: no cycle

This technique detects loops without modifying the list or using extra memory.

✅ Python Solution:

# 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:
        slow = head
        fast = head

        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next

            if slow == fast:
                return True

        return False

⏱ Time & Space Complexity:

• Time: O(n) — each pointer visits each node at most once

• Space: O(1) — constant space, no extra data structures

💬 TL;DR:

• Don’t track nodes in a set.

• Use two pointers.

• If they meet, you found a cycle. If not, no cycle.

This trick shows up again and again in linked list problems.
It’s an interview favorite — master it once, use it forever