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In this post, you will find the solution for the **Linked List Cycle II** **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 II– LeetCode Problem

Linked List Cycle II– LeetCode Problem

**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 = 1Output:tail connects to node index 1Explanation:There is a cycle in the linked list, where tail connects to the second node.

**Example 2:**

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

**Example 3:**

Input:head = [1], pos = -1Output:no cycleExplanation: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 II– LeetCode Solutions

Linked List Cycle II Solution in C++:

class Solution { public: ListNode* detectCycle(ListNode* head) { ListNode* slow = head; ListNode* fast = head; while (fast && fast->next) { slow = slow->next; fast = fast->next->next; if (slow == fast) { slow = head; while (slow != fast) { slow = slow->next; fast = fast->next; } return slow; } } return nullptr; } };

Linked List Cycle II Solution in Java:

public class Solution { public ListNode detectCycle(ListNode head) { ListNode slow = head; ListNode fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) { slow = head; while (slow != fast) { slow = slow.next; fast = fast.next; } return slow; } } return null; } }

Linked List Cycle II Solution in Python:

class Solution: def detectCycle(self, head: ListNode) -> ListNode: slow = head fast = head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: slow = head while slow != fast: slow = slow.next fast = fast.next return slow return None