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Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

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In this post, you will find the solution for the **Linked List Cycle** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Linked List Cycle– LeetCode Problem

Linked List Cycle– LeetCode Problem

**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. **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 = 1Output:trueExplanation:There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

**Example 2:**

Input:head = [1,2], pos = 0Output:trueExplanation:There is a cycle in the linked list, where the tail connects to the 0th node.

**Example 3:**

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

**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.

Linked List Cycle– LeetCode Solutions

Linked List Cycle Solution in C++:

class Solution { public: bool hasCycle(ListNode* head) { ListNode* slow = head; ListNode* fast = head; while (fast && fast->next) { slow = slow->next; fast = fast->next->next; if (slow == fast) return true; } return false; } };

Linked List Cycle Solution in Java:

public class Solution { public boolean hasCycle(ListNode head) { 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; } }

Linked List Cycle Solution in Python:

class Solution: def hasCycle(self, head: 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