Linked List Cycle

Easy Linked List Slow Fast Pointers Original on LeetCode

Linked List Cycle is a easy linked list problem solved with the slow fast pointers pattern. The best approach, optimal (floyd's slow and fast pointers), runs in O(n) time and O(1) space. Below are 2 approaches in Java, from hash set of visited nodes up.

Problem

Given the head of a linked list, decide whether it contains a cycle: some node can be reached again by following next pointers.

Examples

Example 1

Input: head = [3, 2, 0, -4], tail connects to index 1
Output: true

Example 2

Input: head = [1], no cycle
Output: false

Constraints

The list has 0 to 10^4 nodes.
Follow-up: O(1) memory.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Hash set of visited nodes	`O(n)`	`O(n)`
Optimal (Floyd's slow and fast pointers)	`O(n)`	`O(1)`

1Hash set of visited nodes

TimeO(n)

SpaceO(n)

Remember every node visited. Seeing a node twice means there is a cycle.

Walk the list; if the set already has the node, return true; else add it.

Java

class Solution {
    public boolean hasCycle(ListNode head) {
        Set<ListNode> seen = new HashSet<>();
        for (ListNode p = head; p != null; p = p.next)
            if (!seen.add(p)) return true;
        return false;
    }
}

2Optimal (Floyd's slow and fast pointers)

TimeO(n)

SpaceO(1)

Move slow one step and fast two steps. If there is a cycle, fast gains one step per move inside it and must meet slow. If fast reaches null, there is no cycle.

slow = fast = head.
While fast and fast.next: move; if slow == fast, return true.
Return false.

Java

class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) return true;
        }
        return false;
    }
}

Edge cases to test

Empty list
A single node pointing to itself

Hints

Hint 1

On a circular track, a faster runner eventually laps a slower one.

FAQ

What is the best time complexity for Linked List Cycle?

Optimal (Floyd's slow and fast pointers) runs in O(n) time and O(1) extra space.

Which pattern does Linked List Cycle use?

It is a linked list problem that uses the slow fast pointers pattern. Other problems with the same pattern: Middle of the Linked List, Linked List Cycle II.

Is there a brute force solution for Linked List Cycle?

Yes. Hash set of visited nodes takes O(n) time and O(n) space. Remember every node visited.

Which edge cases should I test for Linked List Cycle?

Empty list; A single node pointing to itself.