141. Linked List Cycle 🔁

Difficulty: Easy - Tags: Linked List, Two Pointers

LeetCode Problem Link

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Problem Statement 📜

Given 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 revisited by continuously following the next pointer. Internally, pos is used to denote the index of the node that the tail's next pointer connects to, but pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

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Examples 🌟

🔹 Example 1:

Input:

head = [3,2,0,-4], pos = 1

Output:

true

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:

Input:

head = [1,2], pos = 0

Output:

true

Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.

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🔹 Example 3:

Input:

head = [1], pos = -1

Output:

false

Explanation: There is no cycle in the linked list.

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Constraints ⚙️

• The number of the nodes in the list is in the range [0, 10^4].

• -10^5 <= Node.val <= 10^5.

• pos is -1 or a valid index in the linked list.

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Follow-Up 🔎

Can you solve it using O(1) (i.e., constant) memory?

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Solution 💡

The problem can be solved using the two-pointer technique (Floyd’s Cycle Detection Algorithm):

1. Use two pointers:

   • A slow pointer moves one step at a time.

   • A fast pointer moves two steps at a time.

2. If there is a cycle, the slow and fast pointers will meet at some point inside the cycle.

3. If the fast pointer reaches the end (null), then there is no cycle.

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Java Solution

class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }

        ListNode slow = head;
        ListNode fast = head;

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;

            // If slow meets fast, there is a cycle
            if (slow == fast) {
                return true;
            }
        }

        return false; // No cycle detected
    }
}

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Explanation of the Solution

1. Two Pointers:

   • The slow pointer moves step by step.

   • The fast pointer skips one step to move faster.

2. Cycle Detection:

   • If a cycle exists, the fast pointer will eventually catch up to the slow pointer.

3. Edge Cases:

   • If the list is empty (head == null) or contains only one node (head.next == null), there can be no cycle.

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Time Complexity ⏳

• O(n): Both pointers traverse the list. In the worst case, they traverse each node once.

Space Complexity 💾

• O(1): The solution uses constant memory, satisfying the follow-up constraint.

You can find the full solution here.