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Add two numbers represented by linked lists | Set 1

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  • Difficulty Level : Easy
  • Last Updated : 24 Jun, 2022

Given two numbers represented by two lists, write a function that returns the sum list. The sum list is a list representation of the addition of two input numbers.

Example:

Input: 
List1: 5->6->3 // represents number 563 
List2: 8->4->2 // represents number 842 
Output: 
Resultant list: 1->4->0->5 // represents number 1405 
Explanation: 563 + 842 = 1405

Input: 
List1: 7->5->9->4->6 // represents number 75946
List2: 8->4 // represents number 84
Output: 
Resultant list: 7->6->0->3->0// represents number 76030
Explanation: 75946+84=76030

Approach: Traverse both lists to the end and add preceding zeros in the list with lesser digits. Then call a recursive function on the start nodes of both lists which calls itself for the next nodes of both lists till it gets to the end. This function creates a node for the sum of the current digits and returns the carry. 

Complete Interview Preparation - GFG

The steps are: 

  1. Traverse the two linked lists in order to add preceding zeros in case a list is having lesser digits than the other one.
  2. Start from the head node of both lists and call a recursive function for the next nodes.
  3. Continue it till the end of the lists.
  4. Creates a node for current digits sum and returns the carry.

Below is the implementation of this approach. 

C++




// C++ program to add two numbers
// represented by linked list
#include <bits/stdc++.h>
using namespace std;
  
/* Linked list node */
class Node {
public:
    int data;
    Node* next;
};
  
/* Function to create a
new node with given data */
Node* newNode(int data)
{
    Node* new_node = new Node();
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
  
/* Function to insert a node at the
beginning of the Singly Linked List */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = newNode(new_data);
    /* link the old list off the new node */
    new_node->next = (*head_ref);
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
  
/* Adds contents of two linked lists and
return the head node of resultant list */
Node* addTwoLists(Node* first, Node* second)
{
    // res is head node of the resultant list
    Node* res = NULL;
    Node *temp, *prev = NULL;
    int carry = 0, sum;
  
    // while both lists exist
    while (first != NULL || second != NULL) {
        // Calculate value of next digit in resultant list.
        // The next digit is sum of following things 
          // (i) Carry 
        // (ii) Next digit of first list (if there is a next digit) 
        // (ii) Next digit of second list (if there is a next digit)
        sum = carry + (first ? first->data : 0) + (second ? second->data : 0);
        // update carry for next calculation
        carry = (sum >= 10) ? 1 : 0;
        // update sum if it is greater than 10
        sum = sum % 10;
        // Create a new node with sum as data
        temp = newNode(sum);
        // if this is the first node then set it as head of the resultant list
        if (res == NULL)
            res = temp;
        // If this is not the first node then connect it to the rest.
        else
            prev->next = temp;
        
        // Set prev for next insertion
        prev = temp;
  
        // Move first and second pointers to next nodes
        if (first)
            first = first->next;
        if (second)
            second = second->next;
    }
    if (carry > 0)
        temp->next = newNode(carry);
    // return head of the resultant list
    return res;
}
  
Node* reverse(Node* head)
{
    if (head == NULL || head->next == NULL)
        return head;
    // reverse the rest list and put the first element at the end 
    Node* rest = reverse(head->next);
    head->next->next = head;
    head->next = NULL;
    // fix the head pointer
    return rest;
}
  
// A utility function to print a linked list
void printList(Node* node)
{
    while (node != NULL) {
        cout << node->data << " ";
        node = node->next;
    }
    cout << endl;
}
  
/* Driver code */
int main(void)
{
    Node* res = NULL;
    Node* first = NULL;
    Node* second = NULL;
  
    // create first list 7->5->9->4->6
    push(&first, 6);
    push(&first, 4);
    push(&first, 9);
    push(&first, 5);
    push(&first, 7);
    printf("First List is ");
    printList(first);
  
    // create second list 8->4
    push(&second, 4);
    push(&second, 8);
    cout << "Second List is ";
    printList(second);
  
    // reverse both the lists
    first = reverse(first);
    second = reverse(second);
    // Add the two lists
    res = addTwoLists(first, second);
  
    // reverse the res to get the sum
    res = reverse(res);
    cout << "Resultant list is ";
    printList(res);
    return 0;
}
  
// This code is contributed by Aditya Kumar (adityakumar129)


C




// C program to add two numbers
// represented by linked list
#include <stdio.h>
#include <stdlib.h>
  
/* Linked list node */
typedef struct Node {
    int data;
    struct Node* next;
}Node;
  
/* Function to create a
new node with given data */
Node* newNode(int data)
{
    Node* new_node = (Node *)malloc(sizeof(Node));
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
  
/* Function to insert a node at the
beginning of the Singly Linked List */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = newNode(new_data);
    /* link the old list off the new node */
    new_node->next = (*head_ref);
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
  
/* Adds contents of two linked lists and
return the head node of resultant list */
Node* addTwoLists(Node* first, Node* second)
{
    // res is head node of the resultant list
    Node* res = NULL;
    Node *temp, *prev = NULL;
    int carry = 0, sum;
  
    // while both lists exist
    while (first != NULL || second != NULL) {
        // Calculate value of next digit in resultant list.
        // The next digit is sum of following things 
          // (i) Carry 
        // (ii) Next digit of first list (if there is a next digit) 
        // (ii) Next digit of second list (if there is a next digit)
        sum = carry + (first ? first->data : 0) + (second ? second->data : 0);
        // update carry for next calculation
        carry = (sum >= 10) ? 1 : 0;
        // update sum if it is greater than 10
        sum = sum % 10;
        // Create a new node with sum as data
        temp = newNode(sum);
        // if this is the first node then set it as head of the resultant list
        if (res == NULL)
            res = temp;
        // If this is not the first node then connect it to the rest.
        else
            prev->next = temp;
        
        // Set prev for next insertion
        prev = temp;
  
        // Move first and second pointers to next nodes
        if (first)
            first = first->next;
        if (second)
            second = second->next;
    }
    if (carry > 0)
        temp->next = newNode(carry);
    // return head of the resultant list
    return res;
}
  
Node* reverse(Node* head)
{
    if (head == NULL || head->next == NULL)
        return head;
    // reverse the rest list and put the first element at the end 
    Node* rest = reverse(head->next);
    head->next->next = head;
    head->next = NULL;
    // fix the head pointer
    return rest;
}
  
// A utility function to print a linked list
void printList(Node* node)
{
    while (node != NULL) {
        printf("%d  ",node->data);
        node = node->next;
    }
    printf("\n");
}
  
/* Driver code */
int main(void)
{
    Node* res = NULL;
    Node* first = NULL;
    Node* second = NULL;
  
    // create first list 7->5->9->4->6
    push(&first, 6);
    push(&first, 4);
    push(&first, 9);
    push(&first, 5);
    push(&first, 7);
    printf("First List is ");
    printList(first);
  
    // create second list 8->4
    push(&second, 4);
    push(&second, 8);
    printf("Second List is");
    printList(second);
  
    // reverse both the lists
    first = reverse(first);
    second = reverse(second);
    // Add the two lists
    res = addTwoLists(first, second);
  
    // reverse the res to get the sum
    res = reverse(res);
       printf("Resultant list is ");
    printList(res);
    return 0;
}
  
// This code is contributed by Aditya Kumar (adityakumar129)


Java




// Java program to add two numbers
// represented by linked list
  
class LinkedList {
  
    static Node head1, head2;
  
    static class Node {
  
        int data;
        Node next;
  
        Node(int d) {
            data = d;
            next = null;
        }
    }
  
    /* Adds contents of two linked lists and prints it */
    void addTwoLists(Node first, Node second) {
        Node start1 = new Node(0);
        start1.next = first;
        Node start2 = new Node(0);
        start2.next = second;
  
        addPrecedingZeros(start1, start2);
        Node result = new Node(0);
        if (sumTwoNodes(start1.next, start2.next, result) == 1) {
            Node node = new Node(1);
            node.next = result.next;
            result.next = node;
        }
        printList(result.next);
    }
  
    /* Adds lists and returns the carry */
    private int sumTwoNodes(Node first, Node second, Node result) {
        if (first == null) {
            return 0;
        }
        int number = first.data + second.data + sumTwoNodes(first.next, second.next, result);
        Node node = new Node(number % 10);
        node.next = result.next;
        result.next = node;
        return number / 10;
    }
  
    /* Appends preceding zeros in case a list is having lesser nodes than the other one*/
    private void addPrecedingZeros(Node start1, Node start2) {
        Node next1 = start1.next;
        Node next2 = start2.next;
        while (next1 != null && next2 != null) {
            next1 = next1.next;
            next2 = next2.next;
        }
        if (next1 == null && next2 != null) {
            while (next2 != null) {
                Node node = new Node(0);
                node.next = start1.next;
                start1.next = node;
                next2 = next2.next;
            }
        } else if (next2 == null && next1 != null) {
            while (next1 != null) {
                Node node = new Node(0);
                node.next = start2.next;
                start2.next = node;
                next1 = next1.next;
            }
        }
    }
  
    /* Utility function to print a linked list */
    void printList(Node head) {
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
        System.out.println("");
    }
  
    // Driver Code
    public static void main(String[] args) {
        LinkedList list = new LinkedList();
  
        // creating first list
        list.head1 = new Node(7);
        list.head1.next = new Node(5);
        list.head1.next.next = new Node(9);
        list.head1.next.next.next = new Node(4);
        list.head1.next.next.next.next = new Node(6);
        System.out.print("First List is ");
        list.printList(head1);
  
        // creating second list
        list.head2 = new Node(8);
        list.head2.next = new Node(4);
        System.out.print("Second List is ");
        list.printList(head2);
  
        System.out.print("Resultant List is ");
        // add the two lists and see the result
        list.addTwoLists(head1, head2);
    }
  
    // this code is contributed by *Saurabh321Gupta*
}


C#




// C# program to add two numbers
// represented by linked list
using System;
public class List {
  
      
    public class Node {
  
        public int data;
        public Node next;
  
        public Node(int d) {
            data = d;
            next = null;
        }
    }
    public static  Node head1, head2;
  
  
    /* Adds contents of two linked lists and prints it */
    public void addTwoLists(Node first, Node second) {
        Node start1 = new Node(0);
        start1.next = first;
        Node start2 = new Node(0);
        start2.next = second;
  
        addPrecedingZeros(start1, start2);
        Node result = new Node(0);
        if (sumTwoNodes(start1.next, start2.next, result) == 1) {
            Node node = new Node(1);
            node.next = result.next;
            result.next = node;
        }
        printList(result.next);
    }
  
    /* Adds lists and returns the carry */
    public  int sumTwoNodes(Node first, Node second, Node result) {
        if (first == null) {
            return 0;
        }
        int number = first.data + second.data + sumTwoNodes(first.next, second.next, result);
        Node node = new Node(number % 10);
        node.next = result.next;
        result.next = node;
        return number / 10;
    }
  
    /*
     * Appends preceding zeros in case a list is having lesser nodes than the other
     * one
     */
    public  void addPrecedingZeros(Node start1, Node start2) {
        Node next1 = start1.next;
        Node next2 = start2.next;
        while (next1 != null && next2 != null) {
            next1 = next1.next;
            next2 = next2.next;
        }
        if (next1 == null && next2 != null) {
            while (next2 != null) {
                Node node = new Node(0);
                node.next = start1.next;
                start1.next = node;
                next2 = next2.next;
            }
        } else if (next2 == null && next1 != null) {
            while (next1 != null) {
                Node node = new Node(0);
                node.next = start2.next;
                start2.next = node;
                next1 = next1.next;
            }
        }
    }
  
    /* Utility function to print a linked list */
    public void printList(Node head) {
        while (head != null) {
            Console.Write(head.data + " ");
            head = head.next;
        }
        Console.WriteLine("");
    }
  
    // Driver Code
    public static void Main(String[] args) {
        List list = new List();
  
        // creating first list
        Node head1 = new Node(7);
        head1.next = new Node(5);
        head1.next.next = new Node(9);
        head1.next.next.next = new Node(4);
        head1.next.next.next.next = new Node(6);
        Console.Write("First List is ");
        list.printList(head1);
  
        // creating second list
        Node head2 = new Node(8);
        head2.next = new Node(4);
        Console.Write("Second List is ");
        list.printList(head2);
  
        Console.Write("Resultant List is ");
        // add the two lists and see the result
        list.addTwoLists(head1, head2);
    }
}
  
// This code is contributed by umadevi9616 


Output

First List is 7 5 9 4 6 
Second List is 8 4 
Resultant list is 7 6 0 3 0 

Complexity Analysis: 

  • Time Complexity: O(m + n), where m and n are numbers of nodes in first and second lists respectively. 
    The lists need to be traversed only once.
  • Space Complexity: O(m + n). 
    A temporary linked list is needed to store the output number

Method 2(Using STL): Using the stack data structure

Approach :

  • Create 3 stacks namely s1,s2,s3.
  • Fill s1 with Nodes of list1 and fill s2 with nodes of list2.
  • Fill s3 by creating new nodes and setting the data of new nodes to the sum of s1.top(), s2.top() and carry until list1 and list2 are empty .
  • If the sum >9
    • set carry 1
  • else
    • set carry 0
  • Create a Node(say prev) that will contain the head of the sum List.
  • Link all the elements of s3 from top to bottom
  • return prev
     

Code:

C++




// C++ program to add two numbers represented by Linked
// Lists using Stack
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
    int data;
    Node* next;
};
Node* newnode(int data)
{
    Node* x = new Node();
    x->data = data;
    return x;
}
// function that returns the sum of two numbers represented
// by linked lists
Node* addTwoNumbers(Node* l1, Node* l2)
{
    Node* prev = NULL;
    // Create 3 stacks
    stack<Node*> s1, s2, s3;
    // Fill first stack with first List Elements
    while (l1 != NULL) {
        s1.push(l1);
        l1 = l1->next;
    }
    // Fill second stack with second List Elements
    while (l2 != NULL) {
        s2.push(l2);
        l2 = l2->next;
    }
    int carry = 0;
    // Fill the third stack with the sum of first and second
    // stack
    while (!s1.empty() && !s2.empty()) {
        int sum = s1.top()->data + s2.top()->data + carry;
        Node* temp = newnode(sum % 10);
        s3.push(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s1.pop();
        s2.pop();
    }
    while (!s1.empty()) {
        int sum = carry + s1.top()->data;
        Node* temp = newnode(sum % 10);
        s3.push(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s1.pop();
    }
    while (!s2.empty()) {
        int sum = carry + s2.top()->data;
        Node* temp = newnode(sum % 10);
        s3.push(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s2.pop();
    }
    // If carry is still present create a new node with
    // value 1 and push it to the third stack
    if (carry == 1) {
        Node* temp = newnode(1);
        s3.push(temp);
    }
    // Link all the elements inside third stack with each
    // other
    if (!s3.empty())
        prev = s3.top();
    while (!s3.empty()) {
        Node* temp = s3.top();
        s3.pop();
        if (s3.size() == 0) {
            temp->next = NULL;
        }
        else {
            temp->next = s3.top();
        }
    }
    return prev;
}
// utility functions
// Function that displays the List
void Display(Node* head)
{
    if (head == NULL) {
        return;
    }
    while (head->next != NULL) {
        cout << head->data << " -> ";
        head = head->next;
    }
    cout << head->data << endl;
}
// Function that adds element at the end of the Linked List
void push(Node** head_ref, int d)
{
    Node* new_node = newnode(d);
    new_node->next = NULL;
    if (*head_ref == NULL) {
        new_node->next = *head_ref;
        *head_ref = new_node;
        return;
    }
    Node* last = *head_ref;
    while (last->next != NULL && last != NULL) {
        last = last->next;
    }
    last->next = new_node;
    return;
}
// Driver Program for above Functions
int main()
{
    // Creating two lists
    // first list = 9 -> 5 -> 0
    // second List = 6 -> 7
    Node* first = NULL;
    Node* second = NULL;
    Node* sum = NULL;
    push(&first, 7);
    push(&first, 5);
    push(&first, 9);
    push(&first, 4);
    push(&first, 6);
    push(&second, 8);
    push(&second, 4);
    cout << "First List : ";
    Display(first);
    cout << "Second List : ";
    Display(second);
    sum = addTwoNumbers(first, second);
    cout << "Sum List : ";
    Display(sum);
    return 0;
}


Java




// Java program to add two numbers represented by Linked
// Lists using Stack
import java.util.*;
  
class LinkedList {
  
    static Node head1, head2;
  
    static class Node {
  
        int data;
        Node next;
  
        Node(int d) {
            data = d;
            next = null;
        }
    }
    // function that calculates and prints the sum of two numbers represented
// by linked lists
static void addTwoLists(Node l1, Node l2)
{
    Node prev = null;
    // Create 3 stacks
    Stack<Node> s1 = new Stack<Node>();
    Stack<Node> s2 = new Stack<Node>();
      Stack<Node> s3 = new Stack<Node>();
    // Fill first stack with first List Elements
    while (l1 != null) {
        s1.add(l1);
        l1 = l1.next;
    }
    // Fill second stack with second List Elements
    while (l2 !=null) {
        s2.add(l2);
        l2 = l2.next;
    }
    int carry = 0;
    // Fill the third stack with the sum of first and second
    // stack
    while (!s1.isEmpty() && !s2.isEmpty()) {
        int sum = s1.peek().data + s2.peek().data + carry;
        Node temp = new Node(sum % 10);
        s3.add(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s1.pop();
        s2.pop();
    }
    while (!s1.isEmpty()) {
        int sum = carry + s1.peek().data;
        Node temp = new Node(sum % 10);
        s3.add(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s1.pop();
    }
    while (!s2.isEmpty()) {
        int sum = carry + s2.peek().data;
        Node temp = new Node(sum % 10);
        s3.add(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s2.pop();
    }
    // If carry is still present create a new node with
    // value 1 and push it to the third stack
    if (carry == 1) {
        Node temp = new Node(1);
        s3.add(temp);
    }
    // Link all the elements inside third stack with each
    // other
    if (!s3.isEmpty())
        prev = s3.peek();
    while (!s3.isEmpty()) {
        Node temp = s3.peek();
        s3.pop();
        if (s3.size() == 0) {
            temp.next = null;
        }
        else {
            temp.next = s3.peek();
        }
    }
    printList(prev);
}
    
    /* Utility function to print a linked list */
    static void printList(Node head) {
        while (head != null) {
            System.out.print(head.data + " -> ");
            head = head.next;
        }
        System.out.println("");
    }
  
    // Driver Code
    public static void main(String[] args) {
        LinkedList list = new LinkedList();
  
        // creating first list
        list.head1 = new Node(7);
        list.head1.next = new Node(5);
        list.head1.next.next = new Node(9);
        list.head1.next.next.next = new Node(4);
        list.head1.next.next.next.next = new Node(6);
        System.out.print("First List : ");
        list.printList(head1);
  
        // creating second list
        list.head2 = new Node(8);
        list.head2.next = new Node(4);
        System.out.print("Second List : ");
        list.printList(head2);
  
        System.out.print("Sum List : ");
        // add the two lists and see the result
        list.addTwoLists(head1, head2);
    }
  
    // this code is contributed by Abhijeet Kumar(abhijeet19403)
}


C#




// C# program to add two numbers represented by Linked
// Lists using Stack
using System;
using System.Collections.Generic;
public class List {
  
      
    public class Node {
  
        public int data;
        public Node next;
  
        public Node(int d) {
            data = d;
            next = null;
        }
    }
    public static  Node head1, head2;
// function that calculates and prints the sum of two numbers represented
// by linked lists
    public void addTwoLists(Node l1, Node l2)
{
    Node prev = null;
    // Create 3 stacks
    Stack<Node> s1 = new Stack<Node>();
    Stack<Node> s2 = new Stack<Node>();
      Stack<Node> s3 = new Stack<Node>();
    // Fill first stack with first List Elements
    while (l1 != null) {
        s1.Push(l1);
        l1 = l1.next;
    }
    // Fill second stack with second List Elements
    while (l2 !=null) {
        s2.Push(l2);
        l2 = l2.next;
    }
    int carry = 0;
    // Fill the third stack with the sum of first and second
    // stack
    while (s1.Count!=0 && s2.Count!=0) {
        int sum = s1.Peek().data + s2.Peek().data + carry;
        Node temp = new Node(sum % 10);
        s3.Push(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s1.Pop();
        s2.Pop();
    }
    while (s1.Count!=0) {
        int sum = carry + s1.Peek().data;
        Node temp = new Node(sum % 10);
        s3.Push(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s1.Pop();
    }
    while (s2.Count!=0) {
        int sum = carry + s2.Peek().data;
        Node temp = new Node(sum % 10);
        s3.Push(temp);
        if (sum > 9) {
            carry = 1;
        }
        else {
            carry = 0;
        }
        s2.Pop();
    }
    // If carry is still present create a new node with
    // value 1 and push it to the third stack
    if (carry == 1) {
        Node temp = new Node(1);
        s3.Push(temp);
    }
    // Link all the elements inside third stack with each
    // other
    if (s3.Count!=0)
        prev = s3.Peek();
    while (s3.Count!=0) {
        Node temp = s3.Peek();
        s3.Pop();
        if (s3.Count == 0) {
            temp.next = null;
        }
        else {
            temp.next = s3.Peek();
        }
    }
    printList(prev);
}
    
     
    /* Utility function to print a linked list */
    public void printList(Node head) {
        while (head != null) {
            Console.Write(head.data + " -> ");
            head = head.next;
        }
        Console.WriteLine("");
    }
  
    // Driver Code
    public static void Main(String[] args) {
        List list = new List();
  
        // creating first list
        Node head1 = new Node(7);
        head1.next = new Node(5);
        head1.next.next = new Node(9);
        head1.next.next.next = new Node(4);
        head1.next.next.next.next = new Node(6);
        Console.Write("First List : ");
        list.printList(head1);
  
        // creating second list
        Node head2 = new Node(8);
        head2.next = new Node(4);
        Console.Write("Second List : ");
        list.printList(head2);
  
        Console.Write("Resultant List : ");
        // add the two lists and see the result
        list.addTwoLists(head1, head2);
    }
}
  
// This code is contributed by Abhijeet Kumar(abhijeet19403)


Output

First List : 7 -> 5 -> 9 -> 4 -> 6
Second List : 8 -> 4
Sum List : 7 -> 6 -> 0 -> 3 -> 0

Time Complextiy: O(n)

Here n is length of the larger list 

Auxiliary Space: O(n)

Extra space is used in storing the elements in the stack.

Another Approach with time complexity O(N):

The given approach works as following steps: 

  1. First, we calculate the sizes of both the linked lists, size1 and size2, respectively.
  2. Then we traverse the bigger linked list, if any, and decrement till the size of both become the same.
  3. Now we traverse both linked lists till the end.
  4. Now the backtracking occurs while performing addition.
  5. Finally, the head node is returned to the linked list containing the answer.

C++




#include <iostream>
using namespace std;
  
struct Node {
    int data;
    struct Node* next;
};
  
// recursive function
Node* addition(Node* temp1, Node* temp2, int size1,
               int size2)
{
    // creating a new Node
    Node* newNode
        = (struct Node*)malloc(sizeof(struct Node));
  
    // base case
    if (temp1->next == NULL && temp2->next == NULL) {
        // addition of current nodes which is the last nodes
        // of both linked lists
        newNode->data = (temp1->data + temp2->data);
  
        // set this current node's link null
        newNode->next = NULL;
  
        // return the current node
        return newNode;
    }
  
    // creating a node that contains sum of previously added
    // number
    Node* returnedNode
        = (struct Node*)malloc(sizeof(struct Node));
  
    // if sizes are same then we move in both linked list
    if (size2 == size1) {
        // recursively call the function
        // move ahead in both linked list
        returnedNode = addition(temp1->next, temp2->next,
                                size1 - 1, size2 - 1);
  
        // add the current nodes and append the carry
        newNode->data = (temp1->data + temp2->data)
                        + ((returnedNode->data) / 10);
    }
    // or else we just move in big linked list
    else {
        // recursively call the function
        // move ahead in big linked list
        returnedNode = addition(temp1, temp2->next, size1,
                                size2 - 1);
  
        // add the current node and carry
        newNode->data
            = (temp2->data) + ((returnedNode->data) / 10);
    }
  
    // this node contains previously added numbers
    // so we need to set only rightmost digit of it
    returnedNode->data = (returnedNode->data) % 10;
  
    // set the returned node to the current node
    newNode->next = returnedNode;
  
    // return the current node
    return newNode;
}
  
// Function to add two numbers represented by nexted list.
struct Node* addTwoLists(struct Node* head1,
                         struct Node* head2)
{
    struct Node *temp1, *temp2, *ans = NULL;
  
    temp1 = head1;
    temp2 = head2;
  
    int size1 = 0, size2 = 0;
  
    // calculating the size of first linked list
    while (temp1 != NULL) {
        temp1 = temp1->next;
        size1++;
    }
    // calculating the size of second linked list
    while (temp2 != NULL) {
        temp2 = temp2->next;
        size2++;
    }
  
    Node* returnedNode
        = (struct Node*)malloc(sizeof(struct Node));
  
    // traverse the bigger linked list
    if (size2 > size1) {
        returnedNode = addition(head1, head2, size1, size2);
    }
    else {
        returnedNode = addition(head2, head1, size2, size1);
    }
  
    // creating new node if head node is >10
    if (returnedNode->data >= 10) {
        ans = (struct Node*)malloc(sizeof(struct Node));
        ans->data = (returnedNode->data) / 10;
        returnedNode->data = returnedNode->data % 10;
        ans->next = returnedNode;
    }
    else
        ans = returnedNode;
  
    // return the head node of linked list that contains
    // answer
    return ans;
}
  
void Display(Node* head)
{
    if (head == NULL) {
        return;
    }
    while (head->next != NULL) {
        cout << head->data << " -> ";
        head = head->next;
    }
    cout << head->data << endl;
}
// Function that adds element at the end of the Linked List
void push(Node** head_ref, int d)
{
    Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));
    new_node->data = d;
    new_node->next = NULL;
    if (*head_ref == NULL) {
        new_node->next = *head_ref;
        *head_ref = new_node;
        return;
    }
    Node* last = *head_ref;
    while (last->next != NULL && last != NULL) {
        last = last->next;
    }
    last->next = new_node;
    return;
}
// Driver Program for above Functions
int main()
{
    // Creating two lists
    Node* first = NULL;
    Node* second = NULL;
    Node* sum = NULL;
    push(&first, 7);
    push(&first, 5);
    push(&first, 9);
    push(&first, 4);
    push(&first, 6);
    push(&second, 8);
    push(&second, 4);
    cout << "First List : ";
    Display(first);
    cout << "Second List : ";
    Display(second);
    sum = addTwoLists(first, second);
    cout << "Sum List : ";
    Display(sum);
    return 0;
}
  
// This code is contributed by Dharmik Parmar


Output

First List : 7 -> 5 -> 9 -> 4 -> 6
Second List : 8 -> 4
Sum List : 7 -> 6 -> 0 -> 3 -> 0

Related Article: Add two numbers represented by linked lists | Set 2

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
 

Another Approach with easy to understand code (without appending zeros)- 

In this approach we simulate how in reality we add two numbers. In the code we have taken 9->8->7 and 1->2->3 as two numbers to add. What we do is reverse these two lists to get 7->8->9 and 3->2->1 and start from the head of the lists to add numbers of individual nodes like we would in practice if we add two numbers.

For example,  first we add 7 and 3 to get 10, which means carry = 1 and value of new node will be 0. Now we continue this till the end of the list.

Steps- 

  1. Reverse the two number lists.
  2. Simulate addition on nodes one by one. Append each node before the already calculated sum nodes.( You will better understand this step in code)
  3. In the end we will get the final answer and we can return the head node.

C++




// C++ Code to add two nodes by reversing the two lists
  
#include <bits/stdc++.h>
using namespace std;
  
/* Linked list Node */
struct Node
{
    int data;
    struct Node *next;
    Node(int x)
    {
        data = x;
        next = NULL;
    }
};
  
void printList(Node *n)
{
    while (n)
    {
        cout << n->data << " ";
        n = n->next;
    }
    cout << endl;
}
  
struct Node *reverseList(struct Node *list)
{
    Node *prev = NULL, *cur = list, *next = NULL;
  
    while (cur != NULL)
    {
        next = cur->next;
        cur->next = prev;
        prev = cur;
        cur = next;
    }
  
    return prev;
}
  
//----------------------------------------------------------------------------
  
struct Node *addTwoLists(struct Node *first, struct Node *second)
{
    // code here
    first = reverseList(first);
    second = reverseList(second);
  
    int carry = 0;
    Node *head = NULL, *prev = NULL;
    Node *sum = NULL;
  
    while (first != NULL or second != NULL or carry == 1) //if any one of these is left we are stil left with addition
    {
        int newVal = carry;
  
        if (first) 
            newVal += first->data;
        if (second)
            newVal += second->data;
  
        carry = newVal / 10; //to be used in the next node calculation
        newVal = newVal % 10;
  
        Node *newNode = new Node(newVal);
        newNode->next = sum; //appending in the beginning of the final ans list, this way we do not have to reverse in the end
        sum = newNode;
  
        if (first) // initialising nodes for next iteration
            first = first->next;
        if (second)
            second = second->next;
    }
  
    return sum;
}
  
//----------------------------------------------------------------------------
  
// { Driver Code Starts.
  
int main()
{
    Node *first1 = new Node(9);
    Node *first2 = new Node(8);
    Node *first3 = new Node(7);
  
    first1->next = first2;
    first2->next = first3;
    first3->next = NULL;
  
    Node *second1 = new Node(1);
    Node *second2 = new Node(2);
    Node *second3 = new Node(3);
  
    second1->next = second2;
    second2->next = second3;
    second3->next = NULL;
  
    Node *ans = addTwoLists(first1, second1);
      
      cout << "Sum is : ";
    printList(ans);
  
    return 0;
}


Java




// Java program to add two numbers represented by Linked
// Lists by reversing lists
import java.util.*;
  
class LinkedList {
  
    static Node head1, head2;
  
    static class Node {
  
        int data;
        Node next;
  
        Node(int d) {
            data = d;
            next = null;
        }
    }
  // function to reverse the linked list and return the head of the reversed list
  static Node reverseList(Node list)
{
    Node prev = null, curr = list, next = null;
    while (curr != null)
    {
        next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}
    // function that calculates and prints the sum of two numbers represented
// by linked lists
  
static void addTwoLists(Node first, Node second)
{
    // code here
    first = reverseList(first);
    second = reverseList(second);
  
    int carry = 0;
    Node head = null, prev = null;
    Node sum = null;
  
    while (first != null || second != null || carry == 1) //if any one of these is left we are stil left with addition
    {
        int newVal = carry;
  
        if (first!=null
            newVal += first.data;
        if (second!=null)
            newVal += second.data;
  
        carry = newVal / 10; //to be used in the next node calculation
        newVal = newVal % 10;
  
        Node newNode = new Node(newVal);
        newNode.next = sum; //appending in the beginning of the final ans list, this way we do not have to reverse in the end
        sum = newNode;
  
        if (first!=null) // initialising nodes for next iteration
            first = first.next;
        if (second!=null)
            second = second.next;
    }
  
    printList(sum);
}
    
    /* Utility function to print a linked list */
    static void printList(Node head) {
        while (head != null) {
            System.out.print(head.data + " -> ");
            head = head.next;
        }
        System.out.println("");
    }
  
    // Driver Code
    public static void main(String[] args) {
        LinkedList list = new LinkedList();
  
        // creating first list
        list.head1 = new Node(9);
        list.head1.next = new Node(8);
        list.head1.next.next = new Node(7);
        System.out.print("First List : ");
        list.printList(head1);
        
        // creating second list
        list.head2 = new Node(1);
        list.head2.next = new Node(2);
          list.head2.next.next = new Node(3);
        System.out.print("Second List : ");
        list.printList(head2);
  
        System.out.print("Sum List : ");
        // add the two lists and see the result
        list.addTwoLists(head1, head2);
    }
  
    // this code is contributed by Abhijeet Kumar(abhijeet19403)
}


C#




// C# program to add two numbers represented by Linked
// Lists by reversing lists
using System;
using System.Collections.Generic;
public class List {
  
      
    public class Node {
  
        public int data;
        public Node next;
  
        public Node(int d) {
            data = d;
            next = null;
        }
    }
    public static  Node head1, head2;
    
  // function to reverse the linked list and return the head of the reversed list
  static Node reverseList(Node list)
{
    Node prev = null, curr = list, next = null;
    while (curr != null)
    {
        next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}
    // function that calculates and prints the sum of two numbers represented
// by linked lists
  
    
static void addTwoLists(Node first, Node second)
{
    // code here
    first = reverseList(first);
    second = reverseList(second);
  
    int carry = 0;
   
    Node sum = null;
    //if any one of these is left we are stil left with addition
    while (first != null || second != null || carry == 1) 
    {
        int newVal = carry;
  
        if (first!=null
            newVal += first.data;
        if (second!=null)
            newVal += second.data;
  
        carry = newVal / 10; //to be used in the next node calculation
        newVal = newVal % 10;
  
        Node newNode = new Node(newVal);
        newNode.next = sum; 
          //appending in the beginning of the final ans list,
        //this way we do not have to reverse in the end
        sum = newNode;
  
        if (first!=null) // initialising nodes for next iteration
            first = first.next;
        if (second!=null)
            second = second.next;
    }
  
    printList(sum);
    /* Utility function to print a linked list */
    static void printList(Node head) {
        while (head != null) {
            Console.Write(head.data + " -> ");
            head = head.next;
        }
        Console.WriteLine("");
    }
  
    // Driver Code
    public static void Main(String[] args) {
          
  
        // creating first list
        Node head1 = new Node(9);
        head1.next = new Node(8);
        head1.next.next = new Node(7);
        Console.Write("First List : ");
        printList(head1);
  
        // creating second list
        Node head2 = new Node(1);
        head2.next = new Node(2);
          head2.next.next = new Node(3);
        Console.Write("Second List : ");
        printList(head2);
  
        Console.Write("Resultant List : ");
        // add the two lists and see the result
        addTwoLists(head1, head2);
    }
}
  
// This code is contributed by Abhijeet Kumar(abhijeet19403)


Output

Sum is : 1 1 1 0 

Complexity Analysis:

Time Complexity: O(max(m,n)), where m and n are numbers of nodes in first and second lists respectively. 
The lists need to be traversed only once.
Space Complexity: O(1)
As constant extra space is used.


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