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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 all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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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LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either Easy, Medium, or Hard.

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• Mathematics/Basic Logical Based Questions
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• Greedy Algorithms
• Breadth-First Search
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• BST (Binary Search Tree)
• Database
• Linked List
• Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

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 = 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 = [1], 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.

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