Detect if a Linked List is Circular in DSA Java - Build from Scratch

Detect if a Linked List is Circular

📖 Scenario: Imagine you have a chain of connected train cars. Sometimes, the last car connects back to an earlier car, making a loop. We want to check if the train cars form such a loop.

🎯 Goal: Build a simple linked list and write code to check if it is circular (has a loop).

📋 What You'll Learn

Create a linked list with 4 nodes containing values 10, 20, 30, 40

Add a boolean variable called isCircular to store the result

Write a method isCircularList that returns true if the list is circular, false otherwise

Print the value of isCircular after checking the list

💡 Why This Matters

🌍 Real World

Detecting loops in linked lists is important in software that manages chains of data or tasks, such as scheduling or network routing.

💼 Career

Understanding how to detect circular references helps in debugging and optimizing data structures in software engineering roles.

1

Create a simple linked list

Create a class called Node with an integer data and a Node reference called next. Then create 4 nodes with values 10, 20, 30, and 40. Link them in order so that the first node points to the second, the second to the third, and the third to the fourth. The fourth node's next should be null.

DSA Java

// Define Node class and create 4 linked nodes
// Your code here

Hint

Remember to create a Node class with data and next. Then link the nodes one by one.

2

Add a boolean variable to store circular check result

Inside the main method, create a boolean variable called isCircular and set it to false. This will hold the result of whether the list is circular or not.

DSA Java

class Node {
    int data;
    Node next;
    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

public class Main {
    public static void main(String[] args) {
        Node head = new Node(10);
        Node second = new Node(20);
        Node third = new Node(30);
        Node fourth = new Node(40);
        head.next = second;
        second.next = third;
        third.next = fourth;
        fourth.next = null;

        // Create a boolean variable called isCircular and set it to false
        // Your code here
    }
}

Hint

Just write boolean isCircular = false; inside main.

3

Write method to detect if the list is circular

Add a static method called isCircularList that takes a Node head as input and returns true if the list is circular, otherwise false. Use two pointers: slow and fast. Move slow by one step and fast by two steps. If they meet, the list is circular. If fast or fast.next becomes null, the list is not circular. Then, inside main, set isCircular to the result of calling isCircularList(head).

DSA Java

class Node {
    int data;
    Node next;
    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

public class Main {
    public static boolean isCircularList(Node head) {
        // Your code here
    }

    public static void main(String[] args) {
        Node head = new Node(10);
        Node second = new Node(20);
        Node third = new Node(30);
        Node fourth = new Node(40);
        head.next = second;
        second.next = third;
        third.next = fourth;
        fourth.next = null;

        boolean isCircular = false;

        // Call isCircularList and assign result to isCircular
        // Your code here
    }
}

Hint

Use two pointers moving at different speeds to detect a loop. If they meet, return true.

4

Print the result

Add a print statement inside main to display the value of isCircular. This will show true if the list is circular, or false if it is not.

DSA Java

class Node {
    int data;
    Node next;
    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

public class Main {
    public static boolean isCircularList(Node head) {
        if (head == null) return false;
        Node slow = head;
        Node fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                return true;
            }
        }
        return false;
    }

    public static void main(String[] args) {
        Node head = new Node(10);
        Node second = new Node(20);
        Node third = new Node(30);
        Node fourth = new Node(40);
        head.next = second;
        second.next = third;
        third.next = fourth;
        fourth.next = null;

        boolean isCircular = false;

        isCircular = isCircularList(head);

        // Print the value of isCircular
        // Your code here
    }
}

Hint

Use System.out.println(isCircular); to print the result.

Practice

1. What does it mean if a linked list is circular?

easy

A. The linked list nodes are sorted in ascending order

B. The last node points back to one of the previous nodes, forming a loop

C. The linked list has no nodes

D. The linked list contains only one node

Solution

1. Step 1: Understand linked list structure

   A linked list is a chain of nodes where each node points to the next node.

2. Step 2: Define circular linked list

   If the last node points back to any previous node, it creates a loop, making the list circular.

3. Final Answer:

   The last node points back to one of the previous nodes, forming a loop -> Option B

4. Quick Check:

   Circular linked list = last node points back [OK]

Hint: Circular means last node links back to earlier node [OK]

Common Mistakes:

• Thinking circular means sorted
• Confusing empty list with circular
• Assuming single node list is always circular

2. Which of the following Java code snippets correctly checks if a linked list is circular using two pointers?

easy

A. while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) return true; }

B. while (slow != null) { slow = slow.next; if (slow == fast) return true; }

C. for (int i = 0; i < size; i++) { if (node == null) return false; node = node.next; }

D. if (head == null) return true; else return false;

Solution

1. Step 1: Identify Floyd's cycle detection logic

   Two pointers move at different speeds: slow moves one step, fast moves two steps.

2. Step 2: Check for meeting point

   If slow and fast pointers meet, the list is circular; otherwise, if fast reaches null, it's not circular.

3. Final Answer:

   while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) return true; } -> Option A

4. Quick Check:

   Two pointers meet = circular [OK]

Hint: Use two pointers moving at different speeds [OK]

Common Mistakes:

• Not checking fast.next for null
• Using only one pointer to detect loop
• Returning true when list is empty

3. What will be the output of the following Java code snippet?

class Node { int data; Node next; Node(int d) { data = d; next = null; } }
public class Test {
  public static boolean isCircular(Node head) {
    if (head == null) return false;
    Node slow = head, fast = head.next;
    while (fast != null && fast.next != null) {
      if (slow == fast) return true;
      slow = slow.next;
      fast = fast.next.next;
    }
    return false;
  }
  public static void main(String[] args) {
    Node head = new Node(1);
    head.next = new Node(2);
    head.next.next = new Node(3);
    head.next.next.next = head; // circular link
    System.out.println(isCircular(head));
  }
}

medium

A. Runtime exception

B. false

C. Compilation error

D. true

Solution

1. Step 1: Understand the linked list structure

   The list has nodes 1 -> 2 -> 3, and then 3 points back to 1, making it circular.

2. Step 2: Trace the isCircular method

   Slow starts at 1, fast at 2. They move until they meet because of the loop, so method returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Loop present = true [OK]

Hint: If fast and slow meet, list is circular [OK]

Common Mistakes:

• Starting fast pointer incorrectly
• Not creating circular link properly
• Confusing output with false

4. Identify the bug in this code snippet that tries to detect if a linked list is circular:

public boolean isCircular(Node head) {
  Node slow = head, fast = head;
  while (fast != null && fast.next != null) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow.next == fast.next) return true;
  }
  return false;
}

medium

A. slow pointer should start at head.next

B. fast pointer should move only one step at a time

C. The condition should compare slow and fast, not slow.next and fast.next

D. The loop condition should be while (slow != null)

Solution

1. Step 1: Analyze pointer comparison

   The code compares slow.next and fast.next, but to detect loop, slow and fast themselves must be compared.

2. Step 2: Correct the comparison

   Change if condition to if (slow == fast) to detect when pointers meet inside the loop.

3. Final Answer:

   The condition should compare slow and fast, not slow.next and fast.next -> Option C

4. Quick Check:

   Compare slow and fast pointers directly [OK]

Hint: Compare slow and fast nodes directly to detect loop [OK]

Common Mistakes:

• Comparing next nodes instead of current nodes
• Moving fast pointer incorrectly
• Wrong loop condition

5. Given a linked list where the last node points to the head node, how can you modify the detection method to also return the length of the loop?

hard

A. After detecting the loop with slow == fast, keep slow fixed and move fast until it meets slow again, counting steps

B. Count nodes from head to tail without checking for loop

C. Return the total number of nodes visited before fast reaches null

D. Use a hash set to store nodes and count unique nodes only

Solution

1. Step 1: Detect loop using two pointers

   Use Floyd's cycle detection to find when slow equals fast inside the loop.

2. Step 2: Count loop length

   Keep slow pointer fixed, move fast pointer one step at a time, incrementing count until fast meets slow again.

3. Final Answer:

   After detecting the loop with slow == fast, keep slow fixed and move fast until it meets slow again, counting steps -> Option A

4. Quick Check:

   Count steps in loop after detection [OK]

Hint: Count nodes in loop by moving one pointer until it meets again [OK]

Common Mistakes:

• Counting nodes before loop detection
• Using total nodes instead of loop length
• Not moving pointer correctly to count loop