# Linked List Cycle II – Leetcode Solution

In this post, we are going to solve the 142. Linked List Cycle II problem of Leetcode. This problem 142. Linked List Cycle II is a Leetcode medium level problem. Let’s see the code, 142. Linked List Cycle II – Leetcode Solution.

## Problem

Given the `head` of a linked list, return the node where the cycle begins. If there is no cycle, return `null`.

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 (0-indexed). It is `-1` if there is no cycle. Note that `pos` is not passed as a parameter.

Do not modify the linked list.

### Example 1 :

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

### Example 3 :

``````
Input: head = , pos = -1
Output: no cycle
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 142. Linked List Cycle II – Leetcode Solution.

# Linked List Cycle II – Leetcode Solution

### 142. Linked List Cycle II – 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 slow;
}
}
return null;
}

if(fast == null) return null;
while(slow != fast){
slow = slow.next;
fast = fast.next;
}
return slow;

}
}```

### 142. Linked List Cycle II – Solution in C++

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

### 142. Linked List Cycle II– Solution in Python

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

class Solution:
def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]: