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 = [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.

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

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    
    public ListNode intersectionPointInCycle(ListNode head){
        if(head == null) return null;
        ListNode slow = head;
        ListNode fast = head;
        
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast){
                return slow;
            }
        }
        return null;
    }
    
    public ListNode detectCycle(ListNode head) {
        if(head == null) return null;
        ListNode slow = head;
        ListNode fast = intersectionPointInCycle(head);
        
        if(fast == null) return null;
        while(slow != fast){
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
        
        
    }
}

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(head==NULL)
        {
            return NULL;
        }
        if(head->next==NULL)
        {
            return NULL;
        }
        ListNode* slow=head;
        ListNode* fast=head;
        while(fast->next!=NULL && fast->next->next!=NULL)
        {
            slow=slow->next;
            fast=fast->next->next;
            if(slow==fast)
            {
                ListNode* temp=head;
                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]:
        e=s=f=head
        while f and f.next:
            s=s.next
            f=f.next.next
            if s==f:
                while s!=e:
                    s=s.next
                    e=e.next
                return e
        return None

Note: This problem 142. Linked List Cycle II is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.