### Find First Node of Cycle in a Linked List - The Coding Shala

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## Find First Node of Cycle in a Linked List

**Problem:**

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where the tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: 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 the 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 the 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.

### Find First Node of Cycle in a Linked List Java Program

**Approach:**

Take two-pointer slow and fast. Slow will move one step and fast will two steps. When slow meets fast then cycle exists. Initialize slow pointer to head and then move slow and fast by 1 step. Both slow and fast will meet at the first node of the cycle.

Java Code

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

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