# LeetCode 141. Linked List Cycle

## Description

https://leetcode.com/problems/palindrome-linked-list/

Given a linked list, determine if it has a cycle in it.

To represent a cycle in the given linked list, we use an integer `pos` which represents the position (0-indexed) in the linked list where tail connects to. If `pos` is `-1`, then there is no cycle in the linked list.

Example 1:

```Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.
```

Example 2:

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

Example 3:

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

## Explanation

If any of the node has been visited, the linked list has a cycle.

## Python Solution

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

class Solution:
def hasCycle(self, head: ListNode) -> bool:

visited = set()

while head != None:
if head not in visited:
visited.add(head)
else:
return True

head = head.next

return False
``````
• Time complexity: O(N).
• Space complexity: O(N).

## 2 Thoughts to “LeetCode 141. Linked List Cycle”

1. Vicky says:

/** Java Solution **/

/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {

if(head==null||head.next==null||head.next.next==null)
return false;

ListNode fast=head;
ListNode slow=head;

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

if(fast==slow)
return true;

}
return false;
}
}