Get Length of Linked List in DSA C - 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 nodes are in the list.

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

📋 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 length of the linked list

Write a loop to traverse the linked list and count the nodes

Print the length of the linked list

💡 Why This Matters

🌍 Real World

Linked lists are used in many programs to store data that can grow or shrink. Counting nodes helps understand the size of the data.

💼 Career

Knowing how to work with linked lists and count their length is a basic skill for software developers working with data structures.

1

Create the linked list with 4 nodes

Define a struct called Node with an int data and a Node* next. Then create 4 nodes named node1, node2, node3, and node4 with values 10, 20, 30, and 40 respectively. Link them so that node1.next = &node2, node2.next = &node3, node3.next = &node4, and node4.next = NULL. Finally, create a pointer head that points to node1.

DSA C

#include 

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

Hint

Remember to define the struct before using it. Initialize each node's data and set next to NULL initially. Then link the nodes by setting the next pointers.

2

Create a variable to count the length

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

DSA C

#include 

struct Node {
    int data;
    struct Node* next;
};

int main() {
    struct Node node1 = {10, NULL};
    struct Node node2 = {20, NULL};
    struct Node node3 = {30, NULL};
    struct Node node4 = {40, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = NULL;

    struct Node* head = &node1;

    // Create length variable and set to 0
    // Your code here

    return 0;
}

Hint

Use int length = 0; to start counting from zero.

3

Traverse the linked list and count nodes

Write a while loop that uses a pointer current starting at head. Loop while current != NULL. Inside the loop, increase length by 1 and move current to current->next.

DSA C

#include 

struct Node {
    int data;
    struct Node* next;
};

int main() {
    struct Node node1 = {10, NULL};
    struct Node node2 = {20, NULL};
    struct Node node3 = {30, NULL};
    struct Node node4 = {40, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = NULL;

    struct Node* head = &node1;

    int length = 0;

    // Traverse linked list and count nodes
    // Your code here

    return 0;
}

Hint

Use a pointer current to move through the list. Increase length each time you visit a node.

4

Print the length of the linked list

Use printf to print the text "Length of linked list: " followed by the value of length and a newline.

DSA C

#include 

struct Node {
    int data;
    struct Node* next;
};

int main() {
    struct Node node1 = {10, NULL};
    struct Node node2 = {20, NULL};
    struct Node node3 = {30, NULL};
    struct Node node4 = {40, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = NULL;

    struct Node* head = &node1;

    int length = 0;

    struct Node* current = head;
    while (current != NULL) {
        length++;
        current = current->next;
    }

    // Print the length
    // Your code here

    return 0;
}

Hint

Use printf("Length of linked list: %d\n", length); to show the count.