Search for a Value in Linked List in DSA C - Build from Scratch

Search for a Value in Linked List

📖 Scenario: You are working with a simple linked list that stores numbers. You want to find out if a certain number is present in the list.

🎯 Goal: Build a program that creates a linked list, sets a target number to search for, searches the linked list for that number, and prints whether the number was found or not.

📋 What You'll Learn

Create a linked list with exactly three nodes containing the values 10, 20, and 30 in that order

Create an integer variable called target and set it to 20

Write a function called search that takes the head of the linked list and the target value, and returns 1 if found, 0 otherwise

Print Found if the target is in the list, otherwise print Not Found

💡 Why This Matters

🌍 Real World

Searching in linked lists is a basic operation used in many software systems where data is stored dynamically, such as in memory management or real-time data processing.

💼 Career

Understanding linked list traversal and search is fundamental for software developers working with low-level data structures, embedded systems, or performance-critical applications.

1

Create the linked list with three nodes

Create a struct called Node with an integer data and a pointer to Node called next. Then create three nodes named node1, node2, and node3. Set their data to 10, 20, and 30 respectively. Link node1.next to node2, node2.next to node3, and node3.next to NULL. Finally, create a pointer head that points to node1.

DSA C

# Define the Node struct and create three linked nodes
# Your code here

Hint

Remember to define the struct before main. Then create three variables of type struct Node. Set their data and link their next pointers properly.

2

Set the target value to search

Inside main, create an integer variable called target and set it to 20.

DSA C

#include 
#include 

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

int main() {
    struct Node node1, node2, node3;
    node1.data = 10;
    node1.next = &node2;
    node2.data = 20;
    node2.next = &node3;
    node3.data = 30;
    node3.next = NULL;

    struct Node* head = &node1;

    // Create target variable
    # Your code here

    return 0;
}

Hint

Just declare int target = 20; inside main.

3

Write the search function

Write a function called search that takes two parameters: a pointer to Node called head and an integer target. The function should return 1 if target is found in the linked list, otherwise return 0. Use a while loop to traverse the list.

DSA C

#include 
#include 

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

// Write search function here

int main() {
    struct Node node1, node2, node3;
    node1.data = 10;
    node1.next = &node2;
    node2.data = 20;
    node2.next = &node3;
    node3.data = 30;
    node3.next = NULL;

    struct Node* head = &node1;

    int target = 20;

    // Your code here

    return 0;
}

Hint

Use a while loop to check each node's data. Return 1 if found, else 0 after the loop.

4

Call search and print the result

Inside main, call the search function with head and target. If it returns 1, print Found. Otherwise, print Not Found.

DSA C

#include 
#include 

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

int search(struct Node* head, int target) {
    struct Node* current = head;
    while (current != NULL) {
        if (current->data == target) {
            return 1;
        }
        current = current->next;
    }
    return 0;
}

int main() {
    struct Node node1, node2, node3;
    node1.data = 10;
    node1.next = &node2;
    node2.data = 20;
    node2.next = &node3;
    node3.data = 30;
    node3.next = NULL;

    struct Node* head = &node1;

    int target = 20;

    // Call search and print result
    # Your code here

    return 0;
}

Hint

Use if (search(head, target)) to check the result. Print Found or Not Found accordingly.