# Linked List Cycle – Leetcode Solution

In this post, we are going to solve the 141. Linked List Cycle problem of Leetcode. This problem 141. Linked List Cycle is a Leetcode easy level problem. Let’s see the code, 141. Linked List Cycle – Leetcode Solution.

## 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 = 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 = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.``````

### Example 3 :

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

### Constraints

• The number of the nodes in the list is in the range `[0, 104]`.
• `-105 <= Node.val <= 105`
• `pos` is `-1` or a valid index in the linked-list.

Now, let’s see the code of 141. Linked List Cycle – Leetcode Solution.

# Linked List Cycle – Leetcode Solution

### 141. Linked List Cycle – Solution in Java

```/**
* class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {

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

if (slow == fast)
return true;
}

return false;
}
}```

### 141. Linked List Cycle – Solution in C++

```/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
while(fast!=NULL and fast->next!=NULL){
fast = fast->next->next;
slow = slow->next;
if(fast==slow){return true;}
}
return false;
}
};```

### 141. Linked List Cycle – Solution in Python

```# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
"""
:rtype: bool
"""
try: