Get Length of Linked List in DSA Java - Build from Scratch

Get Length of Linked List

📖 Scenario: You are working with a simple linked list that stores numbers. You want to find out how many numbers are in the list.

🎯 Goal: Build a Java program that creates a linked list, then counts and prints the number of nodes in it.

📋 What You'll Learn

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

Create a variable to count the nodes

Use a loop to go through the linked list and count each node

Print the total count of nodes

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like music playlists, undo features in apps, and managing tasks in order.

💼 Career

Understanding linked lists and how to find their length is a basic skill for software developers working with data structures and memory management.

1

Create the Linked List

Create a class called Node with an int data field and a Node next field. Then create 4 Node objects named head, second, third, and fourth with values 10, 20, 30, and 40 respectively. Link them so that head points to second, second points to third, and third points to fourth. The fourth node should point to null.

DSA Java

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

Hint

Remember to create a Node class with a constructor that sets the data and initializes next to null. Then create each node and link them by setting the next field.

2

Create a Counter Variable

Inside the main method, create an integer variable called count and set it to 0. This will keep track of how many nodes are in the linked list.

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 an integer variable called count and set it to 0
        // Your code here
    }
}

Hint

Just write int count = 0; inside the main method after creating the linked list.

3

Count the Nodes in the Linked List

Use a Node variable called current and set it to head. Then use a while loop that runs as long as current is not null. Inside the loop, increase count by 1 and move current to the next node by setting current = current.next.

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;

        int count = 0;

        // Use a while loop to count nodes
        // Your code here
    }
}

Hint

Start with current = head; then use while (current != null) to loop. Inside the loop, add 1 to count and move current to current.next.

4

Print the Length of the Linked List

Add a System.out.println statement to print the value of count after the loop ends. This will show the total number of nodes in the linked list.

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;

        int count = 0;

        Node current = head;
        while (current != null) {
            count++;
            current = current.next;
        }

        // Print the count
        // Your code here
    }
}

Hint

Use System.out.println(count); to print the number of nodes after counting them.

Practice

1. What does the length of a linked list represent?

easy

A. The sum of all node values

B. The value of the first node

C. The memory size of the linked list

D. The number of nodes in the linked list

Solution

1. Step 1: Understand linked list structure

   A linked list is made of nodes connected one after another.

2. Step 2: Define length meaning

   The length is how many nodes are present in the list.

3. Final Answer:

   The number of nodes in the linked list -> Option D

4. Quick Check:

   Length = count of nodes [OK]

Hint: Length counts nodes, not values or memory [OK]

Common Mistakes:

• Confusing length with sum of values
• Thinking length is memory size
• Using first node value as length

2. Which of the following Java code snippets correctly initializes a variable to count the length of a linked list named head?

easy

A. int length = head; Node current = 0;

B. int length = 0; Node current = head;

C. Node length = 0; int current = head;

D. int length = null; Node current = head;

Solution

1. Step 1: Identify correct variable types

   Length should be an integer starting at 0; current should be a Node starting at head.

2. Step 2: Check each option

   int length = 0; Node current = head; correctly declares int length = 0; and Node current = head;. Others mix types or assign invalid values.

3. Final Answer:

   int length = 0; Node current = head; -> Option B

4. Quick Check:

   Initialize count as int 0 and pointer as Node head [OK]

Hint: Count is int zero; pointer is Node starting at head [OK]

Common Mistakes:

• Mixing variable types
• Assigning head to int
• Using null for int variable

3. What is 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 int getLength(Node head) {
    int count = 0;
    Node current = head;
    while (current != null) {
      count++;
      current = current.next;
    }
    return count;
  }

  public static void main(String[] args) {
    Node n1 = new Node(5);
    Node n2 = new Node(10);
    Node n3 = new Node(15);
    n1.next = n2;
    n2.next = n3;
    System.out.println(getLength(n1));
  }
}

medium

A. 0

B. 2

C. 3

D. NullPointerException

Solution

1. Step 1: Trace linked list nodes

   Nodes are n1 -> n2 -> n3, so total 3 nodes linked.

2. Step 2: Count nodes in getLength

   Loop increments count for each node until current is null, so count = 3.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   3 nodes counted in list [OK]

Hint: Count increments once per node until null [OK]

Common Mistakes:

• Counting nodes incorrectly
• Expecting 0 for non-empty list
• Confusing null pointer errors

4. Identify the error in the following code that tries to get the length of a linked list:

public int getLength(Node head) {
  int count = 0;
  while (head.next != null) {
    count++;
    head = head.next;
  }
  return count;
}

medium

A. It does not count the last node

B. It causes infinite loop

C. It throws NullPointerException immediately

D. It counts nodes twice

Solution

1. Step 1: Analyze loop condition

   The loop runs while head.next != null, so it stops before last node.

2. Step 2: Check counting logic

   Since last node's next is null, it is never counted, so count misses last node.

3. Final Answer:

   It does not count the last node -> Option A

4. Quick Check:

   Loop condition excludes last node [OK]

Hint: Loop on current != null to count all nodes [OK]

Common Mistakes:

• Using head.next != null instead of head != null
• Missing last node count
• Confusing loop conditions

5. Given a linked list where each node contains an integer, write a method to find the length of the list but only count nodes with even values. Which approach is correct?

hard

A. Traverse all nodes, increment count only if node's data % 2 == 0

B. Count nodes until first odd value is found

C. Count all nodes and then subtract odd nodes count

D. Only count nodes if next node is even

Solution

1. Step 1: Understand problem requirement

   We must count nodes with even values only, ignoring odd nodes.

2. Step 2: Choose correct counting method

   Traverse entire list, check each node's data; increment count if even (data % 2 == 0).

3. Final Answer:

   Traverse all nodes, increment count only if node's data % 2 == 0 -> Option A

4. Quick Check:

   Count even nodes by checking each node [OK]

Hint: Check each node's value; count only if even [OK]

Common Mistakes:

• Stopping count at first odd node
• Subtracting odd count after full count
• Checking next node's value instead of current