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 exists in the list.

🎯 Goal: Build a linked list with given numbers, then write code to search for a specific number in the list and print whether it is found or not.

📋 What You'll Learn

Create a linked list with nodes containing the exact values: 10, 20, 30, 40, 50

Create a variable called target with the value 30

Write a loop to search for target in the linked list

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

💡 Why This Matters

🌍 Real World

Searching in linked lists is common in many programs where data is stored in a chain of elements, like playlists, task lists, or navigation paths.

💼 Career

Understanding how to traverse and search linked lists is a fundamental skill for software developers working with data structures and memory management.

1

Create the linked list

Create a linked list with nodes containing these exact values in order: 10, 20, 30, 40, 50. Use a struct called Node with an int data and a Node* next. Create a pointer called head that points to the first node.

DSA C++

// Define the Node struct and create the linked list with values 10, 20, 30, 40, 50
// Your code here

Hint

Start by defining the Node struct with data and next. Then create nodes one by one and link them using next.

2

Set the target value to search

Create an integer variable called target and set it to 30.

DSA C++

#include 
using namespace std;

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

int main() {
    Node* head = new Node{10, nullptr};
    head->next = new Node{20, nullptr};
    head->next->next = new Node{30, nullptr};
    head->next->next->next = new Node{40, nullptr};
    head->next->next->next->next = new Node{50, nullptr};

    // Create an int variable called target and set it to 30
    // Your code here

    return 0;
}

Hint

Just write int target = 30; inside main().

3

Search for the target value in the linked list

Write a while loop to go through the linked list starting from head. Use a pointer called current to move through nodes. Inside the loop, check if current->data equals target. If yes, break the loop. If not, move current to the next node.

DSA C++

#include 
using namespace std;

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

int main() {
    Node* head = new Node{10, nullptr};
    head->next = new Node{20, nullptr};
    head->next->next = new Node{30, nullptr};
    head->next->next->next = new Node{40, nullptr};
    head->next->next->next->next = new Node{50, nullptr};

    int target = 30;

    // Use a while loop with a pointer current to search for target
    // Your code here

    return 0;
}

Hint

Use Node* current = head; to start. Then loop while current != nullptr. Inside, check current->data == target. If yes, break;. Else, move current = current->next;.

4

Print if the target was found or not

After the loop, check if current is not nullptr. If it is not nullptr, print Found. Otherwise, print Not Found.

DSA C++

#include 
using namespace std;

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

int main() {
    Node* head = new Node{10, nullptr};
    head->next = new Node{20, nullptr};
    head->next->next = new Node{30, nullptr};
    head->next->next->next = new Node{40, nullptr};
    head->next->next->next->next = new Node{50, nullptr};

    int target = 30;

    Node* current = head;
    while (current != nullptr) {
        if (current->data == target) {
            break;
        }
        current = current->next;
    }

    // Print Found if current is not nullptr, else print Not Found
    // Your code here

    return 0;
}

Hint

Use if (current != nullptr) to check if found. Print Found or Not Found accordingly.

Practice

1. What is the main purpose of searching for a value in a linked list?

easy

A. To add a new node at the end of the list

B. To find if a specific value exists in the list

C. To delete the first node of the list

D. To reverse the order of nodes in the list

Solution

1. Step 1: Understand the search operation

   Searching in a linked list means checking each node to see if it contains the value we want.

2. Step 2: Identify the goal of searching

   The goal is to find whether the value exists anywhere in the list, not to modify the list.

3. Final Answer:

   To find if a specific value exists in the list -> Option B

4. Quick Check:

   Search means find value presence = To find if a specific value exists in the list [OK]

Hint: Search means check each node for the value [OK]

Common Mistakes:

• Confusing search with insertion or deletion
• Thinking search changes the list
• Assuming search reverses the list

2. Which of the following is the correct syntax to start traversing a singly linked list in C++?

easy

A. Node* current = new Node();

B. Node current = head;

C. Node current = new Node();

D. Node* current = head;

Solution

1. Step 1: Understand pointer usage in linked lists

   In C++, linked list nodes are accessed via pointers, so we use a pointer variable to traverse.

2. Step 2: Identify correct pointer initialization

   We set a pointer 'current' to point to the head node to start traversal: Node* current = head;

3. Final Answer:

   Node* current = head; -> Option D

4. Quick Check:

   Use pointer to start traversal = Node* current = head; [OK]

Hint: Use pointer to traverse linked list starting at head [OK]

Common Mistakes:

• Using object instead of pointer for traversal
• Creating a new node instead of using existing head
• Not initializing traversal pointer correctly

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

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

Node* head = new Node{10, nullptr};
head->next = new Node{20, nullptr};
head->next->next = new Node{30, nullptr};

Node* current = head;
int searchValue = 20;
bool found = false;
while (current != nullptr) {
  if (current->data == searchValue) {
    found = true;
    break;
  }
  current = current->next;
}
if (found) std::cout << "Value found";
else std::cout << "Value not found";

medium

A. Value not found

B. Compilation error

C. Value found

D. Runtime error

Solution

1. Step 1: Trace the linked list nodes and values

   The list has nodes with values 10 -> 20 -> 30. We search for 20.

2. Step 2: Check the search loop logic

   The loop checks each node's data. When it reaches node with 20, found becomes true and loop breaks.

3. Final Answer:

   Value found -> Option C

4. Quick Check:

   20 exists in list, so output = "Value found" [OK]

Hint: Check each node's data until match or end [OK]

Common Mistakes:

• Forgetting to update current pointer
• Using wrong comparison operator
• Not breaking loop after finding value

4. Identify the error in the following code that searches for a value in a linked list:

Node* current = head;
int searchValue = 15;
while (current->next != nullptr) {
  if (current->data == searchValue) {
    std::cout << "Found";
    break;
  }
  current = current->next;
}
if (current == nullptr) std::cout << "Not Found";

medium

A. The loop condition skips checking the last node

B. The searchValue variable is not initialized

C. The current pointer is not updated inside the loop

D. The output statements are missing semicolons

Solution

1. Step 1: Analyze the loop condition

   The loop runs while current->next is not nullptr, so it stops before the last node.

2. Step 2: Understand impact on search

   The last node's data is never checked, so if the value is there, it will be missed.

3. Final Answer:

   The loop condition skips checking the last node -> Option A

4. Quick Check:

   Loop must check all nodes including last [OK]

Hint: Loop on current != nullptr to check all nodes [OK]

Common Mistakes:

• Using current->next != nullptr instead of current != nullptr
• Not checking last node's data
• Incorrectly handling nullptr checks

5. You want to search for a value in a singly linked list and return its position (1-based index). Which approach correctly implements this?

int searchPosition(Node* head, int value) {
  int pos = 1;
  Node* current = head;
  while (current != nullptr) {
    if (current->data == value) return pos;
    current = current->next;
    pos++;
  }
  return -1;
}

hard

A. Returns the position of value or -1 if not found

B. Returns true if value found, false otherwise

C. Returns the last node's data always

D. Returns 0 if value is found at head

Solution

1. Step 1: Understand the function logic

   The function starts position at 1 and traverses nodes, checking data against value.

2. Step 2: Check return values

   If value matches, it returns current position; if not found, returns -1.

3. Final Answer:

   Returns the position of value or -1 if not found -> Option A

4. Quick Check:

   Function returns index or -1 if missing [OK]

Hint: Return index on match, else -1 after full traversal [OK]

Common Mistakes:

• Returning boolean instead of position
• Starting position count at 0 instead of 1
• Not returning -1 when value not found