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Linked List Cycle - LeetCode Daily Challenge

Prerna Sharma

Problem Statement

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. Note that pos is not passed as a parameter.

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

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).

Example 2

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Try here before watching the video.

Video Solution

Java Code

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

C++ Code

class Solution {
    bool 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 == fast)
                return true;
        return false;

Python Code

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if head is None or is None:
            return False

        slow = head
        fast = head
        while fast is not None and is not None:
            slow =
            fast =
            if slow == fast:
                return True
        return False

Javascript Code

var hasCycle = function(head) {
    if (head === null || === null)
        return false;

    let slow = head;
    let fast = head;
    while (fast !== null && !== null) {
        slow =;
        fast =;
        if (slow === fast)
            return true;
    return false;

Go Code

func hasCycle(head *ListNode) bool {
    if head == nil || head.Next == nil {
        return false

    slow := head
    fast := head
    for fast != nil && fast.Next != nil {
        slow = slow.Next
        fast = fast.Next.Next
        if slow == fast {
            return true
    return false

Complexity Analysis

Time Complexity: O(N), where N is the number of nodes in the linked list.
Space Complexity: O(1), we are not utilizing any extra space.