142. Linked List Cycle II LeetCode Solution

In this guide, you will get 142. Linked List Cycle II LeetCode Solution with the best time and space complexity. The solution to Linked List Cycle II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Linked List Cycle II solution in C++
4. Linked List Cycle II solution in Java
5. Linked List Cycle II solution in Python
6. Additional Resources

Problem Statement of Linked List Cycle II

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.

Follow up: Can you solve it using O(1) (i.e. constant) memory?

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(1)

142. Linked List Cycle II LeetCode Solution in C++

class Solution {
 public:
  ListNode* detectCycle(ListNode* head) {
    ListNode* slow = head;
    ListNode* fast = head;

    while (fast != nullptr && fast->next != nullptr) {
      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;
  }
};
/* code provided by PROGIEZ */

142. Linked List Cycle II LeetCode Solution in Java

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;
  }
}
// code provided by PROGIEZ

142. Linked List Cycle II LeetCode 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
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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