Add Two Numbers Represented as Linked List in DSA C++ - Build from Scratch

Add Two Numbers Represented as Linked List

📖 Scenario: You are working on a simple calculator app that adds two numbers. These numbers are stored in reverse order as linked lists, where each node contains a single digit.For example, the number 342 is stored as 2 -> 4 -> 3 -> null.

🎯 Goal: Build a program that adds two numbers represented as linked lists and returns the sum as a linked list in the same reversed order.

📋 What You'll Learn

Create two linked lists representing the numbers 342 and 465 respectively.

Create a helper variable to track the carry during addition.

Write the logic to add the two linked lists digit by digit.

Print the resulting linked list representing the sum.

💡 Why This Matters

🌍 Real World

Adding numbers digit by digit is similar to how calculators or computers handle large number arithmetic internally.

💼 Career

Understanding linked lists and how to manipulate them is important for software development roles that involve data structures and algorithmic problem solving.

1

Create the linked lists for the two numbers

Create a linked list called l1 representing the number 342 as 2 -> 4 -> 3 -> nullptr and another linked list called l2 representing the number 465 as 5 -> 6 -> 4 -> nullptr. Use the provided ListNode struct.

DSA C++

#include 

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

int main() {

    // Create linked list l1: 2 -> 4 -> 3 -> nullptr
    // Create linked list l2: 5 -> 6 -> 4 -> nullptr

    // Your code here

    return 0;
}

Hint

Use the new keyword to create nodes and link them using the next pointer.

2

Create a carry variable for addition

Create an integer variable called carry and set it to 0 to keep track of the carry during addition.

DSA C++

#include 

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

int main() {
    ListNode* l1 = new ListNode(2);
    l1->next = new ListNode(4);
    l1->next->next = new ListNode(3);

    ListNode* l2 = new ListNode(5);
    l2->next = new ListNode(6);
    l2->next->next = new ListNode(4);

    // Create carry variable
    // Your code here

    return 0;
}

Hint

Initialize carry to zero before starting the addition.

3

Add the two linked lists digit by digit

Write a while loop that continues as long as l1, l2, or carry is not zero. Inside the loop, add the values of l1 and l2 nodes and the carry. Create new nodes for the result linked list with the digit value and update carry accordingly.

DSA C++

#include 

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

int main() {
    ListNode* l1 = new ListNode(2);
    l1->next = new ListNode(4);
    l1->next->next = new ListNode(3);

    ListNode* l2 = new ListNode(5);
    l2->next = new ListNode(6);
    l2->next->next = new ListNode(4);

    int carry = 0;

    // Add l1 and l2 digit by digit
    // Your code here

    return 0;
}

Hint

Use a dummy head node to simplify list construction. Update carry and move pointers forward.

4

Print the resulting linked list

Print the values of the resulting linked list starting from dummyHead->next in the format 7 -> 0 -> 8 -> nullptr.

DSA C++

#include 

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

int main() {
    ListNode* l1 = new ListNode(2);
    l1->next = new ListNode(4);
    l1->next->next = new ListNode(3);

    ListNode* l2 = new ListNode(5);
    l2->next = new ListNode(6);
    l2->next->next = new ListNode(4);

    int carry = 0;

    ListNode* dummyHead = new ListNode(0);
    ListNode* current = dummyHead;

    while (l1 != nullptr || l2 != nullptr || carry != 0) {
        int val1 = (l1 != nullptr) ? l1->val : 0;
        int val2 = (l2 != nullptr) ? l2->val : 0;
        int sum = val1 + val2 + carry;
        carry = sum / 10;
        current->next = new ListNode(sum % 10);
        current = current->next;
        if (l1 != nullptr) l1 = l1->next;
        if (l2 != nullptr) l2 = l2->next;
    }

    // Print the resulting linked list
    // Your code here

    return 0;
}

Hint

Traverse the result linked list and print each node's value followed by ->. End with nullptr.

Practice

1. What does the problem "Add Two Numbers Represented as Linked List" mainly involve?

easy

A. Adding two numbers where each digit is stored in a node of a linked list

B. Sorting two linked lists in ascending order

C. Finding the middle node of a linked list

D. Reversing a linked list

Solution

1. Step 1: Understand the problem context

   The problem involves two linked lists where each node contains a single digit of a number, stored in reverse order.

2. Step 2: Identify the required operation

   We need to add these two numbers digit by digit, handling carry, and return the sum as a linked list.

3. Final Answer:

   Adding two numbers where each digit is stored in a node of a linked list -> Option A

4. Quick Check:

   Adding digits in linked lists = Adding two numbers where each digit is stored in a node of a linked list [OK]

Hint: Think of linked list nodes as digits of a number [OK]

Common Mistakes:

• Confusing addition with sorting
• Thinking the problem is about reversing lists
• Assuming the digits are stored in normal order

2. Which of the following C++ code snippets correctly declares a node structure for a singly linked list used in this problem (using nullptr)?

easy

A. struct Node { int val; Node* next; Node(int x) : val(x), next(NULL) {} };

B. struct Node { int val; Node next; Node(int x) : val(x), next(nullptr) {} };

C. struct Node { int val; Node* next; Node(int x) : val(x), next(nullptr) {} };

D. struct Node { int val; Node* next; Node(int x) { val = x; next = NULL; } };

Solution

1. Step 1: Check pointer usage in struct

   Node* next is correct to point to the next node; struct Node { int val; Node next; Node(int x) : val(x), next(nullptr) {} }; uses Node next which is incorrect (causes infinite recursion).

2. Step 2: Check constructor initialization

   struct Node { int val; Node* next; Node(int x) : val(x), next(nullptr) {} }; uses member initializer list correctly with nullptr; struct Node { int val; Node* next; Node(int x) { val = x; next = NULL; } }; uses assignment inside constructor body which is valid but less modern; struct Node { int val; Node* next; Node(int x) : val(x), next(NULL) {} }; uses NULL which is legacy.

3. Final Answer:

   struct Node { int val; Node* next; Node(int x) : val(x), next(nullptr) {} }; -> Option C

4. Quick Check:

   Use pointer and nullptr in constructor [OK]

Hint: Use Node* next and initialize with nullptr [OK]

Common Mistakes:

• Using Node next instead of Node* next
• Using NULL instead of nullptr
• Not initializing next pointer

3. What is the output linked list after adding the two numbers represented by these linked lists? List1: 2 -> 4 -> 3 -> null (represents 342) List2: 5 -> 6 -> 4 -> null (represents 465) Choose the correct resulting linked list:

medium

A. 7 -> 10 -> 7 -> null

B. 7 -> 0 -> 8 -> null

C. 8 -> 0 -> 7 -> null

D. 7 -> 1 -> 8 -> null

Solution

1. Step 1: Add digits node by node with carry

   2+5=7 (carry 0), 4+6=10 (carry 1), 3+4+1=8 (carry 0)

2. Step 2: Form the resulting linked list

   Digits are 7, 0 (from 10 carry), 8 so list is 7 -> 0 -> 8 -> null

3. Final Answer:

   7 -> 0 -> 8 -> null -> Option B

4. Quick Check:

   342 + 465 = 807 represented as 7->0->8 [OK]

Hint: Add digits with carry, store remainder in node [OK]

Common Mistakes:

• Not handling carry correctly
• Adding digits in wrong order
• Confusing node values with actual numbers

4. Consider the following code snippet for adding two numbers represented as linked lists. What is the error?

Node* addTwoNumbers(Node* l1, Node* l2) {
  Node* dummy = new Node(0);
  Node* current = dummy;
  int carry = 0;
  while (l1 != nullptr && l2 != nullptr) {
    int sum = l1->val + l2->val + carry;
    carry = sum / 10;
    current->next = new Node(sum % 10);
    current = current->next;
    l1 = l1->next;
    l2 = l2->next;
  }
  return dummy->next;
}

medium

A. The loop condition should be while (l1 != nullptr || l2 != nullptr)

B. Carry is not updated correctly

C. Dummy node should be deleted before return

D. The sum calculation is incorrect

Solution

1. Step 1: Analyze loop condition

   The loop runs only while both l1 and l2 are not null, so it stops early if one list is longer.

2. Step 2: Correct loop condition

   It should continue while either l1 or l2 has nodes to process all digits, so use || instead of &&.

3. Final Answer:

   The loop condition should be while (l1 != nullptr || l2 != nullptr) -> Option A

4. Quick Check:

   Use || to handle different length lists [OK]

Hint: Use || in loop to cover all nodes [OK]

Common Mistakes:

• Using && causes missing nodes in longer list
• Ignoring carry after loop ends
• Not advancing pointers correctly

5. Given two linked lists representing numbers in reverse order, how would you modify the addition algorithm to handle the case when the final carry is non-zero after processing all nodes? Choose the best approach:

hard

A. Ignore the carry since it is not needed

B. Subtract the carry from the last node value

C. Restart the addition from the beginning

D. Add a new node with the carry value at the end of the result list

Solution

1. Step 1: Understand carry after full addition

   After adding all nodes, if carry is still 1, it means the sum has an extra digit.

2. Step 2: Add new node for carry

   To represent this extra digit, create a new node with carry value and link it at the end.

3. Final Answer:

   Add a new node with the carry value at the end of the result list -> Option D

4. Quick Check:

   Carry leftover means add new node [OK]

Hint: Add carry node if leftover after loop ends [OK]

Common Mistakes:

• Ignoring leftover carry causing wrong sum
• Trying to subtract carry instead of adding node
• Restarting addition unnecessarily