Get Length of Linked List in DSA Java

Mental Model

Count how many nodes are in the linked list by moving from the start to the end one by one.

Analogy: Imagine counting people standing in a line by starting at the first person and moving down the line until no one is left.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null

Goal: Find the total number of nodes in the linked list

Step 1: start at head node with value 1

[curr->1] -> 2 -> 3 -> null

Why: We begin counting from the first node

Step 2: move curr to next node with value 2, count becomes 1

1 -> [curr->2] -> 3 -> null

Why: Count each node as we move forward

Step 3: move curr to next node with value 3, count becomes 2

1 -> 2 -> [curr->3] -> null

Why: Continue counting nodes until the end

Step 4: move curr to null, stop counting

1 -> 2 -> 3 -> [curr->null]

Why: Reached end of list, counting complete

Result:

1 -> 2 -> 3 -> null
Length = 3

Annotated Code

DSA Java

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

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

    public static void main(String[] args) {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);

        int length = getLength(head);
        System.out.println("Length = " + length);
    }
}

while (curr != null) {

loop through each node until end of list

count++;

increment count for each node visited

curr = curr.next;

advance curr pointer to next node

Output

Length = 3

Complexity Analysis

Time: O(n) because we visit each node once to count

Space: O(1) because we use only a few variables regardless of list size

vs Alternative: Alternative methods like recursion use extra space on the call stack, so this iterative approach is more space efficient

Edge Cases

empty list (head is null)

count remains zero and function returns 0

DSA Java

while (curr != null) {

single node list

count becomes 1 after one iteration

DSA Java

while (curr != null) {

When to Use This Pattern

When asked to find the size or length of a linked list, use a simple traversal counting nodes until you reach the end.

Common Mistakes

Mistake: Forgetting to move the current pointer to the next node inside the loop
Fix: Add 'curr = curr.next;' inside the while loop to advance the pointer

Mistake: Initializing count to 1 instead of 0, causing off-by-one errors
Fix: Initialize count to 0 before starting the loop

Summary

Counts the number of nodes in a linked list by traversing it once.

Use when you need to know how many elements are in a linked list.

The key is to move through each node one by one and count until you reach the end.

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