Add two numbers represented by linked lists | Set 2

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Given two numbers represented by two linked lists, write a function that returns the sum of the two linked lists in the form of a list.

Note: It is not allowed to modify the lists. Also, not allowed to use explicit extra space (Hint: Use Recursion).

Example :

Input: First List: 5->6->3, Second List: 8->4->2
Output: Resultant list: 1->4->0->5
Explanation: Sum of 563 and 842 is 1405

We have discussed a solution here which is for linked lists where the least significant digit is the first node of lists and the most significant digit is the last node. In this problem, the most significant digit is the first node and the least significant digit is the last node and we are not allowed to modify the lists. Recursion is used here to calculate the sum from right to left.

Following are the steps.
1) Calculate sizes of given two linked lists.
2) If sizes are same, then calculate sum using recursion. Hold all nodes in recursion call stack till the rightmost node, calculate the sum of rightmost nodes and forward carry to the left side.
3) If size is not same, then follow below steps:
….a) Calculate difference of sizes of two linked lists. Let the difference be diff
….b) Move diff nodes ahead in the bigger linked list. Now use step 2 to calculate the sum of the smaller list and right sub-list (of the same size) of a larger list. Also, store the carry of this sum.
….c) Calculate the sum of the carry (calculated in the previous step) with the remaining left sub-list of a larger list. Nodes of this sum are added at the beginning of the sum list obtained the previous step.

Below is a dry run of the above approach:

Below is the implementation of the above approach.

C++

// A C++ recursive program to add two linked lists
#include <bits/stdc++.h>
using namespace std;
 
// A linked List Node
class Node {
public:
    int data;
    Node* next;
};
 
typedef Node node;
 
/* A utility function to insert
a node at the beginning of linked list */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = new Node[(sizeof(Node))];
 
    /* put in the data */
    new_node->data = new_data;
 
    /* link the old list of the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* A utility function to print linked list */
void printList(Node* node)
{
    while (node != NULL) {
        cout << node->data << " ";
        node = node->next;
    }
    cout << endl;
}
 
// A utility function to swap two pointers
void swapPointer(Node** a, Node** b)
{
    node* t = *a;
    *a = *b;
    *b = t;
}
 
/* A utility function to get size of linked list */
int getSize(Node* node)
{
    int size = 0;
    while (node != NULL) {
        node = node->next;
        size++;
    }
    return size;
}
 
// Adds two linked lists of same size
// represented by head1 and head2 and returns
// head of the resultant linked list. Carry
// is propagated while returning from the recursion
node* addSameSize(Node* head1, Node* head2, int* carry)
{
    // Since the function assumes linked lists are of same
    // size, check any of the two head pointers
    if (head1 == NULL)
        return NULL;
 
    int sum;
 
    // Allocate memory for sum node of current two nodes
    Node* result = new Node[(sizeof(Node))];
 
    // Recursively add remaining nodes and get the carry
    result->next
        = addSameSize(head1->next, head2->next, carry);
 
    // add digits of current nodes and propagated carry
    sum = head1->data + head2->data + *carry;
    *carry = sum / 10;
    sum = sum % 10;
 
    // Assign the sum to current node of resultant list
    result->data = sum;
 
    return result;
}
 
// This function is called after the
// smaller list is added to the bigger
// lists's sublist of same size. Once the
// right sublist is added, the carry
// must be added toe left side of larger
// list to get the final result.
void addCarryToRemaining(Node* head1, Node* cur, int* carry,
                         Node** result)
{
    int sum;
 
    // If diff. number of nodes are not traversed, add carry
    if (head1 != cur) {
        addCarryToRemaining(head1->next, cur, carry,
                            result);
 
        sum = head1->data + *carry;
        *carry = sum / 10;
        sum %= 10;
 
        // add this node to the front of the result
        push(result, sum);
    }
}
 
// The main function that adds two linked lists
// represented by head1 and head2. The sum of
// two lists is stored in a list referred by result
void addList(Node* head1, Node* head2, Node** result)
{
    Node* cur;
 
    // first list is empty
    if (head1 == NULL) {
        *result = head2;
        return;
    }
 
    // second list is empty
    else if (head2 == NULL) {
        *result = head1;
        return;
    }
 
    int size1 = getSize(head1);
    int size2 = getSize(head2);
 
    int carry = 0;
 
    // Add same size lists
    if (size1 == size2)
        *result = addSameSize(head1, head2, &carry);
 
    else {
        int diff = abs(size1 - size2);
 
        // First list should always be larger than second
        // list. If not, swap pointers
        if (size1 < size2)
            swapPointer(&head1, &head2);
 
        // move diff. number of nodes in first list
        for (cur = head1; diff--; cur = cur->next)
            ;
 
        // get addition of same size lists
        *result = addSameSize(cur, head2, &carry);
 
        // get addition of remaining first list and carry
        addCarryToRemaining(head1, cur, &carry, result);
    }
 
    // if some carry is still there, add a new node to the
    // front of the result list. e.g. 999 and 87
    if (carry)
        push(result, carry);
}
 
// Driver code
int main()
{
    Node *head1 = NULL, *head2 = NULL, *result = NULL;
 
    int arr1[] = { 9, 9, 9 };
    int arr2[] = { 1, 8 };
 
    int size1 = sizeof(arr1) / sizeof(arr1[0]);
    int size2 = sizeof(arr2) / sizeof(arr2[0]);
 
    // Create first list as 9->9->9
    int i;
    for (i = size1 - 1; i >= 0; --i)
        push(&head1, arr1[i]);
 
    // Create second list as 1->8
    for (i = size2 - 1; i >= 0; --i)
        push(&head2, arr2[i]);
 
    addList(head1, head2, &result);
 
    printList(result);
 
    return 0;
}
 
// This code is contributed by rathbhupendra

C

// A C recursive program to add two linked lists
 
#include <stdio.h>
#include <stdlib.h>
 
// A linked List Node
struct Node {
    int data;
    struct Node* next;
};
 
typedef struct Node node;
 
/* A utility function to insert a 
  node at the beginning of
 * linked list */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));
 
    /* put in the data  */
    new_node->data = new_data;
 
    /* link the old list of the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* A utility function to print linked list */
void printList(struct Node* node)
{
    while (node != NULL) {
        printf("%d  ", node->data);
        node = node->next;
    }
    printf("n");
}
 
// A utility function to swap two pointers
void swapPointer(Node** a, Node** b)
{
    node* t = *a;
    *a = *b;
    *b = t;
}
 
/* A utility function to get size 
   of linked list */
int getSize(struct Node* node)
{
    int size = 0;
    while (node != NULL) {
        node = node->next;
        size++;
    }
    return size;
}
 
// Adds two linked lists of same 
// size represented by head1
// and head2 and returns head of 
// the resultant linked list.
// Carry is propagated while 
// returning from the recursion
node* addSameSize(Node* head1,
                  Node* head2, int* carry)
{
    // Since the function assumes
    // linked lists are of same
    // size, check any of the two 
    // head pointers
    if (head1 == NULL)
        return NULL;
 
    int sum;
 
    // Allocate memory for sum 
    // node of current two nodes
    Node* result = (Node*)malloc(sizeof(Node));
 
    // Recursively add remaining nodes
    // and get the carry
    result->next
        = addSameSize(head1->next, 
                      head2->next, carry);
 
    // add digits of current nodes 
    // and propagated carry
    sum = head1->data + head2->data + *carry;
    *carry = sum / 10;
    sum = sum % 10;
 
    // Assigne the sum to current 
    // node of resultant list
    result->data = sum;
 
    return result;
}
 
// This function is called after 
// the smaller list is added
// to the bigger lists's sublist 
// of same size.  Once the
// right sublist is added, the 
// carry must be added toe left
// side of larger list to get 
// the final result.
void addCarryToRemaining(Node* head1, 
                         Node* cur, int* carry,
                         Node** result)
{
    int sum;
 
    // If diff. number of nodes are 
    // not traversed, add carry
    if (head1 != cur) {
        addCarryToRemaining(head1->next,
                            cur, carry,
                            result);
 
        sum = head1->data + *carry;
        *carry = sum / 10;
        sum %= 10;
 
        // add this node to the front of the result
        push(result, sum);
    }
}
 
// The main function that adds two 
// linked lists represented
// by head1 and head2. The sum of
// two lists is stored in a
// list referred by result
void addList(Node* head1, 
             Node* head2, Node** result)
{
    Node* cur;
 
    // first list is empty
    if (head1 == NULL) {
        *result = head2;
        return;
    }
 
    // second list is empty
    else if (head2 == NULL)
    {
        *result = head1;
        return;
    }
 
    int size1 = getSize(head1);
    int size2 = getSize(head2);
 
    int carry = 0;
 
    // Add same size lists
    if (size1 == size2)
        *result = addSameSize(head1, head2, &carry);
 
    else {
        int diff = abs(size1 - size2);
 
        // First list should always be 
        // larger than second
        // list. If not, swap pointers
        if (size1 < size2)
            swapPointer(&head1, &head2);
 
        // move diff. number of nodes in first list
        for (cur = head1; diff--; cur = cur->next)
            ;
 
        // get addition of same size lists
        *result = addSameSize(cur, 
                              head2, &carry);
 
        // get addition of remaining first list and carry
        addCarryToRemaining(head1, 
                            cur, &carry, result);
    }
 
    // if some carry is still there, add a new node to the
    // front of the result list. e.g. 999 and 87
    if (carry)
        push(result, carry);
}
 
// Driver code
int main()
{
    Node *head1 = NULL, *head2 = NULL, *result = NULL;
 
    int arr1[] = { 9, 9, 9 };
    int arr2[] = { 1, 8 };
 
    int size1 = sizeof(arr1) / sizeof(arr1[0]);
    int size2 = sizeof(arr2) / sizeof(arr2[0]);
 
    // Create first list as 9->9->9
    int i;
    for (i = size1 - 1; i >= 0; --i)
        push(&head1, arr1[i]);
 
    // Create second list as 1->8
    for (i = size2 - 1; i >= 0; --i)
        push(&head2, arr2[i]);
 
    addList(head1, head2, &result);
 
    printList(result);
 
    return 0;
}

Java

// A Java recursive program to add two linked lists
 
public class linkedlistATN 
{
    class node 
    {
        int val;
        node next;
 
        public node(int val) 
        {
            this.val = val;
        }
    }
     
    // Function to print linked list
    void printlist(node head) 
    {
        while (head != null) 
        {
            System.out.print(head.val + " ");
            head = head.next;
        }
    }
 
    node head1, head2, result;
    int carry;
 
    /* A utility function to push a value to linked list */
    void push(int val, int list) 
    {
        node newnode = new node(val);
        if (list == 1) 
        {
            newnode.next = head1;
            head1 = newnode;
        } 
        else if (list == 2) 
        {
            newnode.next = head2;
            head2 = newnode;
        } 
        else
        {
            newnode.next = result;
            result = newnode;
        }
 
    }
 
    // Adds two linked lists of same size represented by
    // head1 and head2 and returns head of the resultant 
    // linked list. Carry is propagated while returning 
    // from the recursion
    void addsamesize(node n, node m) 
    {
        // Since the function assumes linked lists are of 
        // same size, check any of the two head pointers
        if (n == null)
            return;
 
        // Recursively add remaining nodes and get the carry
        addsamesize(n.next, m.next);
 
        // add digits of current nodes and propagated carry
        int sum = n.val + m.val + carry;
        carry = sum / 10;
        sum = sum % 10;
 
        // Push this to result list
        push(sum, 3);
 
    }
 
    node cur;
 
    // This function is called after the smaller list is 
    // added to the bigger lists's sublist of same size. 
    // Once the right sublist is added, the carry must be 
    // added to the left side of larger list to get the 
    // final result.
    void propogatecarry(node head1) 
    {
        // If diff. number of nodes are not traversed, add carry
        if (head1 != cur) 
        {
            propogatecarry(head1.next);
            int sum = carry + head1.val;
            carry = sum / 10;
            sum %= 10;
 
            // add this node to the front of the result
            push(sum, 3);
        }
    }
 
    int getsize(node head) 
    {
        int count = 0;
        while (head != null) 
        {
            count++;
            head = head.next;
        }
        return count;
    }
 
    // The main function that adds two linked lists 
    // represented by head1 and head2. The sum of two 
    // lists is stored in a list referred by result
    void addlists() 
    {
        // first list is empty
        if (head1 == null) 
        {
            result = head2;
            return;
        }
 
        // first list is empty
        if (head2 == null) 
        {
            result = head1;
            return;
        }
 
        int size1 = getsize(head1);
        int size2 = getsize(head2);
 
        // Add same size lists
        if (size1 == size2) 
        {
            addsamesize(head1, head2);
        } 
        else
        {
            // First list should always be larger than second list.
            // If not, swap pointers
            if (size1 < size2) 
            {
                node temp = head1;
                head1 = head2;
                head2 = temp;
            }
            int diff = Math.abs(size1 - size2);
 
            // move diff. number of nodes in first list
            node temp = head1;
            while (diff-- >= 0) 
            {
                cur = temp;
                temp = temp.next;
            }
 
            // get addition of same size lists
            addsamesize(cur, head2);
 
            // get addition of remaining first list and carry
            propogatecarry(head1);
        }
            // if some carry is still there, add a new node to 
            // the front of the result list. e.g. 999 and 87
            if (carry > 0)
                push(carry, 3);
         
    }
 
    // Driver program to test above functions
    public static void main(String args[])
    {
        linkedlistATN list = new linkedlistATN();
        list.head1 = null;
        list.head2 = null;
        list.result = null;
        list.carry = 0;
        int arr1[] = { 9, 9, 9 };
        int arr2[] = { 1, 8 };
 
        // Create first list as 9->9->9
        for (int i = arr1.length - 1; i >= 0; --i)
            list.push(arr1[i], 1);
 
        // Create second list as 1->8
        for (int i = arr2.length - 1; i >= 0; --i)
            list.push(arr2[i], 2);
 
        list.addlists();
 
        list.printlist(list.result);
    }
}
 
// This code is contributed by Rishabh Mahrsee

Python3

# A Python3 recursive program to add two linked lists
class node:
    def __init__(self, val):
        self.val = val
        self.next = None
 
head1, head2, result = None, None, None
carry = 0
 
# Function to print linked list
def printlist(head):
    while head != None:
        print(head.val, end=" ")
        head = head.next
 
#  A utility function to push a value to linked list
def push(val, lst):
    global head1, head2, result
    newnode = node(val)
    if lst == 1:
        newnode.next = head1
        head1 = newnode
    elif lst == 2:
        newnode.next = head2
        head2 = newnode
    else:
        newnode.next = result
        result = newnode
 
# Adds two linked lists of same size represented by
# head1 and head2 and returns head of the resultant
# linked list. Carry is propagated while returning
# from the recursion
def addsamesize(n, m):
    global carry
     
    # Since the function assumes linked lists are of
    # same size, check any of the two head pointers
    if n == None:
        return
       
    # Recursively add remaining nodes and get the carry
    addsamesize(n.next, m.next)
     
    # add digits of current nodes and propagated carry
    sum = n.val + m.val + carry
    carry = int(sum / 10)
    sum = sum % 10
     
    # Push this to result list
    push(sum, 3)
 
cur = None
 
# This function is called after the smaller list is
# added to the bigger lists's sublist of same size.
# Once the right sublist is added, the carry must be
# added to the left side of larger list to get the
# final result.
def propogatecarry(head1):
    global carry, cur
     
    # If diff. number of nodes are not traversed, add carry
    if head1 != cur:
        propogatecarry(head1.next)
        sum = carry + head1.val
        carry = int(sum / 10)
        sum %= 10
         
        # add this node to the front of the result
        push(sum, 3)
 
def getsize(head):
    count = 0
    while head != None:
        count += 1
        head = head.next
    return count
   
# The main function that adds two linked lists
# represented by head1 and head2. The sum of two
# lists is stored in a list referred by result
def addlists():
    global head1, head2, result, carry, cur
     
    # first list is empty
    if head1 == None:
        result = head2
        return
       
    # first list is empty
    if head2 == None:
        result = head1
        return
    size1 = getsize(head1)
    size2 = getsize(head2)
     
    # Add same size lists
    if size1 == size2:
        addsamesize(head1, head2)
    else:
       
        # First list should always be larger than second list.
        # If not, swap pointers
        if size1 < size2:
            temp = head1
            head1 = head2
            head2 = temp
        diff = abs(size1 - size2)
         
        # move diff. number of nodes in first list
        temp = head1
        while diff >= 0:
            cur = temp
            temp = temp.next
            diff -= 1
             
        # get addition of same size lists
        addsamesize(cur, head2)
         
        # get addition of remaining first list and carry
        propogatecarry(head1)
         
    # if some carry is still there, add a new node to
    # the front of the result list. e.g. 999 and 87
    if carry > 0:
        push(carry, 3)
 
# Driver program to test above functions
head1, head2, result = None, None, None
carry = 0
arr1 = [9, 9, 9]
arr2 = [1, 8]
 
# Create first list as 9->9->9
for i in range(len(arr1)-1, -1, -1):
    push(arr1[i], 1)
 
# Create second list as 1->8
for i in range(len(arr2)-1, -1, -1):
    push(arr2[i], 2)
 
addlists()
printlist(result)
 
# This code is contributed by Prajwal Kandekar

C#

// A C# recursive program to add two linked lists
using System;
  
public class linkedlistATN{
     
class node 
{
    public int val;
    public node next;
 
    public node(int val) 
    {
        this.val = val;
    }
}
  
// Function to print linked list
void printlist(node head) 
{
    while (head != null) 
    {
        Console.Write(head.val + " ");
        head = head.next;
    }
}
 
node head1, head2, result;
int carry;
 
// A utility function to push a 
// value to linked list 
void push(int val, int list) 
{
    node newnode = new node(val);
     
    if (list == 1) 
    {
        newnode.next = head1;
        head1 = newnode;
    } 
    else if (list == 2) 
    {
        newnode.next = head2;
        head2 = newnode;
    } 
    else
    {
        newnode.next = result;
        result = newnode;
    }
 
}
 
// Adds two linked lists of same size represented by
// head1 and head2 and returns head of the resultant 
// linked list. Carry is propagated while returning 
// from the recursion
void addsamesize(node n, node m) 
{
     
    // Since the function assumes linked
    // lists are of same size, check any
    // of the two head pointers
    if (n == null)
        return;
 
    // Recursively add remaining nodes
    // and get the carry
    addsamesize(n.next, m.next);
 
    // Add digits of current nodes 
    // and propagated carry
    int sum = n.val + m.val + carry;
    carry = sum / 10;
    sum = sum % 10;
 
    // Push this to result list
    push(sum, 3);
}
 
node cur;
 
// This function is called after the smaller 
// list is added to the bigger lists's sublist 
// of same size. Once the right sublist is added,
// the carry must be added to the left side of 
// larger list to get the final result.
void propogatecarry(node head1) 
{
     
    // If diff. number of nodes are 
    // not traversed, add carry
    if (head1 != cur) 
    {
        propogatecarry(head1.next);
        int sum = carry + head1.val;
        carry = sum / 10;
        sum %= 10;
 
        // Add this node to the front 
        // of the result
        push(sum, 3);
    }
}
 
int getsize(node head) 
{
    int count = 0;
    while (head != null) 
    {
        count++;
        head = head.next;
    }
    return count;
}
 
// The main function that adds two linked 
// lists represented by head1 and head2. 
// The sum of two lists is stored in a 
// list referred by result
void addlists() 
{
     
    // First list is empty
    if (head1 == null) 
    {
        result = head2;
        return;
    }
 
    // Second list is empty
    if (head2 == null) 
    {
        result = head1;
        return;
    }
 
    int size1 = getsize(head1);
    int size2 = getsize(head2);
 
    // Add same size lists
    if (size1 == size2) 
    {
        addsamesize(head1, head2);
    } 
    else
    {
         
        // First list should always be
        // larger than second list.
        // If not, swap pointers
        if (size1 < size2) 
        {
            node temp = head1;
            head1 = head2;
            head2 = temp;
        }
         
        int diff = Math.Abs(size1 - size2);
 
        // Move diff. number of nodes in
        // first list
        node tmp = head1;
         
        while (diff-- >= 0) 
        {
            cur = tmp;
            tmp = tmp.next;
        }
 
        // Get addition of same size lists
        addsamesize(cur, head2);
 
        // Get addition of remaining 
        // first list and carry
        propogatecarry(head1);
    }
        // If some carry is still there, 
        // add a new node to the front of
        // the result list. e.g. 999 and 87
        if (carry > 0)
            push(carry, 3);
}
 
// Driver code
public static void Main(string []args)
{
    linkedlistATN list = new linkedlistATN();
    list.head1 = null;
    list.head2 = null;
    list.result = null;
    list.carry = 0;
     
    int []arr1 = { 9, 9, 9 };
    int []arr2 = { 1, 8 };
 
    // Create first list as 9->9->9
    for(int i = arr1.Length - 1; i >= 0; --i)
        list.push(arr1[i], 1);
 
    // Create second list as 1->8
    for(int i = arr2.Length - 1; i >= 0; --i)
        list.push(arr2[i], 2);
 
    list.addlists();
 
    list.printlist(list.result);
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
// A javascript recursive program to add two linked lists
 
  class node {
        constructor(val) {
            this.val = val;
            this.next = null;
        }
    }
 
    // Function to print linked list
    function printlist( head) {
        while (head != null) {
            document.write(head.val + " ");
            head = head.next;
        }
    }
 
    var head1, head2, result;
    var carry;
 
    /* A utility function to push a value to linked list */
    function push(val , list) {
        var newnode = new node(val);
        if (list == 1) {
            newnode.next = head1;
            head1 = newnode;
        } else if (list == 2) {
            newnode.next = head2;
            head2 = newnode;
        } else {
            newnode.next = result;
            result = newnode;
        }
 
    }
 
    // Adds two linked lists of same size represented by
    // head1 and head2 and returns head of the resultant
    // linked list. Carry is propagated while returning
    // from the recursion
    function addsamesize( n,  m) {
        // Since the function assumes linked lists are of
        // same size, check any of the two head pointers
        if (n == null)
            return;
 
        // Recursively add remaining nodes and get the carry
        addsamesize(n.next, m.next);
 
        // add digits of current nodes and propagated carry
        var sum = n.val + m.val + carry;
        carry = parseInt(sum / 10);
        sum = sum % 10;
 
        // Push this to result list
        push(sum, 3);
 
    }
 
    var cur;
 
    // This function is called after the smaller list is
    // added to the bigger lists's sublist of same size.
    // Once the right sublist is added, the carry must be
    // added to the left side of larger list to get the
    // final result.
    function propogatecarry( head1) {
        // If diff. number of nodes are not traversed, add carry
        if (head1 != cur) {
            propogatecarry(head1.next);
            var sum = carry + head1.val;
            carry = parseInt(sum / 10);
            sum %= 10;
 
            // add this node to the front of the result
            push(sum, 3);
        }
    }
 
    function getsize( head) {
        var count = 0;
        while (head != null) {
            count++;
            head = head.next;
        }
        return count;
    }
 
    // The main function that adds two linked lists
    // represented by head1 and head2. The sum of two
    // lists is stored in a list referred by result
    function addlists() {
        // first list is empty
        if (head1 == null) {
            result = head2;
            return;
        }
 
        // first list is empty
        if (head2 == null) {
            result = head1;
            return;
        }
 
        var size1 = getsize(head1);
        var size2 = getsize(head2);
 
        // Add same size lists
        if (size1 == size2) {
            addsamesize(head1, head2);
        } else {
            // First list should always be larger than second list.
            // If not, swap pointers
            if (size1 < size2) {
                var temp = head1;
                head1 = head2;
                head2 = temp;
            }
            var diff = Math.abs(size1 - size2);
 
            // move diff. number of nodes in first list
            var temp = head1;
            while (diff-- >= 0) {
                cur = temp;
                temp = temp.next;
            }
 
            // get addition of same size lists
            addsamesize(cur, head2);
 
            // get addition of remaining first list and carry
            propogatecarry(head1);
        }
        // if some carry is still there, add a new node to
        // the front of the result list. e.g. 999 and 87
        if (carry > 0)
            push(carry, 3);
 
    }
 
    // Driver program to test above functions
     
        head1 = null;
        head2 = null;
        result = null;
        carry = 0;
        var arr1 = [ 9, 9, 9 ];
        var arr2 = [ 1, 8 ];
 
        // Create first list as 9->9->9
        for (i = arr1.length - 1; i >= 0; --i)
            push(arr1[i], 1);
 
        // Create second list as 1->8
        for (i = arr2.length - 1; i >= 0; --i)
            push(arr2[i], 2);
 
        addlists();
 
        printlist(result);
 
// This code is contributed by todaysgaurav
</script>

Output

1 0 1 7

Time Complexity: O(m+n) where m and n are the sizes of given two linked lists.
Auxiliary Space: O(m+n) for call stack

Iterative Approach:

This implementation does not have any recursion call overhead, which means it is an iterative solution.

Since we need to start adding numbers from the last of the two linked lists. So, here we will use the stack data structure to implement this.

We will firstly make two stacks from the given two linked lists.
Then, we will run a loop till both stack become empty.
in every iteration, we keep the track of the carry.
In the end, if carry>0, that means we need extra node at the start of the resultant list to accommodate this carry.

Below is the implementation of the above approach.

C++

// C++ Iterative program to add two linked lists  
#include <bits/stdc++.h> 
using namespace std; 
   
// A linked List Node  
class Node  
{  
    public: 
    int data;  
    Node* next;  
};
 
// to push a new node to linked list
void push(Node** head_ref, int new_data)  
{  
    /* allocate node */
    Node* new_node = new Node[(sizeof(Node))];  
   
    /* put in the data */
    new_node->data = new_data;  
   
    /* link the old list of the new node */
    new_node->next = (*head_ref);  
   
    /* move the head to point to the new node */
    (*head_ref) = new_node;  
}
 
// to add two new numbers
Node* addTwoNumList(Node* l1, Node* l2) {
    stack<int> s1,s2;
    while(l1!=NULL){
        s1.push(l1->data);
        l1=l1->next;
    }
    while(l2!=NULL){
        s2.push(l2->data);
        l2=l2->next;
    }
    int carry=0;
    Node* result=NULL;
    while(s1.empty()==false || s2.empty()==false){
        int a=0,b=0;
        if(s1.empty()==false){
            a=s1.top();s1.pop();
        }
        if(s2.empty()==false){
            b=s2.top();s2.pop();
        }
        int total=a+b+carry;
        Node* temp=new Node();
        temp->data=total%10;
        carry=total/10;
        if(result==NULL){
            result=temp;
        }else{
            temp->next=result;
            result=temp;
        }
    }
    if(carry!=0){
        Node* temp=new Node();
        temp->data=carry;
        temp->next=result;
        result=temp;
    }
    return result;
}
 
// to print a linked list
void printList(Node *node)  
{  
    while (node != NULL)  
    {  
        cout<<node->data<<" ";  
        node = node->next;  
    }  
    cout<<endl;  
}
 
// Driver Code
int main()  
{  
    Node *head1 = NULL, *head2 = NULL;  
   
    int arr1[] = {5, 6, 7};  
    int arr2[] = {1, 8};  
   
    int size1 = sizeof(arr1) / sizeof(arr1[0]);  
    int size2 = sizeof(arr2) / sizeof(arr2[0]);  
   
    // Create first list as 5->6->7  
    int i;  
    for (i = size1-1; i >= 0; --i)  
        push(&head1, arr1[i]);  
   
    // Create second list as 1->8  
    for (i = size2-1; i >= 0; --i)  
        push(&head2, arr2[i]);  
     
    Node* result=addTwoNumList(head1, head2);
    printList(result);  
   
    return 0;  
}

Java

// Java Iterative program to add 
// two linked lists  
import java.io.*;
import java.util.*;
 
class GFG{
     
static class Node
{
    int data;
    Node next;
     
    public Node(int data)
    { 
        this.data = data;
    }
}
 
static Node l1, l2, result;
 
// To push a new node to linked list
public static void push(int new_data)
{
     
    // Allocate node 
    Node new_node = new Node(0);
 
    // Put in the data 
    new_node.data = new_data;
 
    // Link the old list of the new node 
    new_node.next = l1;
 
    // Move the head to point to the new node 
    l1 = new_node;
}
 
public static void push1(int new_data)
{
     
    // Allocate node 
    Node new_node = new Node(0);
 
    // Put in the data 
    new_node.data = new_data;
 
    // Link the old list of the new node 
    new_node.next = l2;
 
    // Move the head to point to
    // the new node 
    l2 = new_node;
}
 
// To add two new numbers
public static Node addTwoNumbers()
{
    Stack<Integer> stack1 = new Stack<>();
    Stack<Integer> stack2 = new Stack<>();
 
    while (l1 != null)
    {
        stack1.add(l1.data);
        l1 = l1.next;
    }
 
    while (l2 != null) 
    {
        stack2.add(l2.data);
        l2 = l2.next;
    }
 
    int carry = 0;
    Node result = null;
 
    while (!stack1.isEmpty() ||
           !stack2.isEmpty()) 
    {
        int a = 0, b = 0;
 
        if (!stack1.isEmpty()) 
        {
            a = stack1.pop();
        }
 
        if (!stack2.isEmpty()) 
        {
            b = stack2.pop();
        }
 
        int total = a + b + carry;
 
        Node temp = new Node(total % 10);
        carry = total / 10;
 
        if (result == null) 
        {
            result = temp;
        }
        else
        {
            temp.next = result;
            result = temp;
        }
    }
 
    if (carry != 0)
    {
        Node temp = new Node(carry);
        temp.next = result;
        result = temp;
    }
    return result;
}
 
// To print a linked list
public static void printList()
{
    while (result != null) 
    {
        System.out.print(result.data + " ");
        result = result.next;
    }
    System.out.println();
}
 
// Driver code
public static void main(String[] args)
{
    int arr1[] = { 5, 6, 7 };
    int arr2[] = { 1, 8 };
 
    int size1 = 3;
    int size2 = 2;
 
    // Create first list as 5->6->7
    int i;
    for(i = size1 - 1; i >= 0; --i)
        push(arr1[i]);
 
    // Create second list as 1->8
    for(i = size2 - 1; i >= 0; --i)
        push1(arr2[i]);
 
    result = addTwoNumbers();
 
    printList();
}
}
 
// This code is contributed by RohitOberoi

Python3

# Python Iterative program to add
# two linked lists    
class Node:
    def __init__(self,val):
        self.data = val
        self.next = None
     
l1, l2, result = None,None,0
 
# To push a new node to linked list
def push(new_data):
 
    global l1
 
    # Allocate node
    new_node = Node(0)
 
    # Put in the data
    new_node.data = new_data
 
    # Link the old list of the new node
    new_node.next = l1
 
    # Move the head to point to the new node
    l1 = new_node
 
 
def push1(new_data):
 
    global l2
 
    # Allocate node
    new_node = Node(0)
 
    # Put in the data
    new_node.data = new_data
 
    # Link the old list of the new node
    new_node.next = l2
 
    # Move the head to point to
    # the new node
    l2 = new_node
 
# To add two new numbers
def addTwoNumbers():
 
    global l1,l2,result
 
    stack1 = []
    stack2 = []
 
    while (l1 != None):
        stack1.append(l1.data)
        l1 = l1.next
 
    while (l2 != None):
        stack2.append(l2.data)
        l2 = l2.next
 
    carry = 0
    result = None
 
    while (len(stack1) != 0 or len(stack2) != 0):
        a,b = 0,0
 
        if (len(stack1) != 0):
            a = stack1.pop()
 
        if (len(stack2) != 0):
            b = stack2.pop()
 
        total = a + b + carry
 
        temp = Node(total % 10)
        carry = total // 10
 
        if (result == None):
            result = temp
        else:
            temp.next = result
            result = temp
 
 
    if (carry != 0):
        temp = Node(carry)
        temp.next = result
        result = temp
         
    return result
 
 
# To print a linked list
def printList():
 
    global result
 
    while (result != None):
        print(result.data ,end = " ")
        result = result.next
 
# Driver code
     
arr1 = [ 5, 6, 7 ]
arr2 = [ 1, 8 ]
 
size1 = 3
size2 = 2
 
# Create first list as 5->6->7
 
for i in range(size1-1,-1,-1):
    push(arr1[i])
 
# Create second list as 1->8
for i in range(size2-1,-1,-1):
    push1(arr2[i])
 
result = addTwoNumbers()
 
printList()
 
# This code is contributed by shinjanpatra

C#

// C# Iterative program to add 
// two linked lists  
using System;
using System.Collections;
 
class GFG{
 
  public class Node
  {
    public int data;
    public Node next;
 
    public Node(int data)
    { 
      this.data = data;
    }
  }
 
  static Node l1, l2, result;
 
  // To push a new node to linked list
  public static void push(int new_data)
  {
 
    // Allocate node 
    Node new_node = new Node(0);
 
    // Put in the data 
    new_node.data = new_data;
 
    // Link the old list of the new node 
    new_node.next = l1;
 
    // Move the head to point to the new node 
    l1 = new_node;
  }
 
  public static void push1(int new_data)
  {
 
    // Allocate node 
    Node new_node = new Node(0);
 
    // Put in the data 
    new_node.data = new_data;
 
    // Link the old list of the new node 
    new_node.next = l2;
 
    // Move the head to point to
    // the new node 
    l2 = new_node;
  }
 
  // To add two new numbers
  public static Node addTwoNumbers()
  {
    Stack stack1 = new Stack();
    Stack stack2 = new Stack();
 
    while (l1 != null)
    {
      stack1.Push(l1.data);
      l1 = l1.next;
    }
    while (l2 != null) 
    {
      stack2.Push(l2.data);
      l2 = l2.next;
    }
 
    int carry = 0;
    Node result = null;
    while (stack1.Count != 0 ||
           stack2.Count != 0) 
    {
      int a = 0, b = 0;
 
      if (stack1.Count != 0) 
      {
        a = (int)stack1.Pop();
      }
 
      if (stack2.Count != 0) 
      {
        b = (int)stack2.Pop();
      }
 
      int total = a + b + carry;
      Node temp = new Node(total % 10);
      carry = total / 10;
 
      if (result == null) 
      {
        result = temp;
      }
      else
      {
        temp.next = result;
        result = temp;
      }
    }
 
    if (carry != 0)
    {
      Node temp = new Node(carry);
      temp.next = result;
      result = temp;
    }
    return result;
  }
 
  // To print a linked list
  public static void printList()
  {
    while (result != null) 
    {
      Console.Write(result.data + " ");
      result = result.next;
    }
    Console.WriteLine();
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int []arr1 = { 5, 6, 7 };
    int []arr2 = { 1, 8 };
    int size1 = 3;
    int size2 = 2;
 
    // Create first list as 5->6->7
    int i;
    for(i = size1 - 1; i >= 0; --i)
      push(arr1[i]);
 
    // Create second list as 1->8
    for(i = size2 - 1; i >= 0; --i)
      push1(arr2[i]);
    result = addTwoNumbers();
    printList();
  }
}
 
// This code is contributed by pratham76

Javascript

<script>
// javascript Iterative program to add 
// two linked lists       
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
    var l1, l2, result;
 
    // To push a new node to linked list
    function push(new_data) {
 
        // Allocate node
var new_node = new Node(0);
 
        // Put in the data
        new_node.data = new_data;
 
        // Link the old list of the new node
        new_node.next = l1;
 
        // Move the head to point to the new node
        l1 = new_node;
    }
 
    function push1(new_data) {
 
        // Allocate node
var new_node = new Node(0);
 
        // Put in the data
        new_node.data = new_data;
 
        // Link the old list of the new node
        new_node.next = l2;
 
        // Move the head to point to
        // the new node
        l2 = new_node;
    }
 
    // To add two new numbers
    function addTwoNumbers() {
        var stack1 = [];
        var stack2 = [];
 
        while (l1 != null) {
            stack1.push(l1.data);
            l1 = l1.next;
        }
 
        while (l2 != null) {
            stack2.push(l2.data);
            l2 = l2.next;
        }
 
        var carry = 0;
var result = null;
 
        while (stack1.length != 0 || stack2.length != 0) {
            var a = 0, b = 0;
 
            if (stack1.length != 0) {
                a = stack1.pop();
            }
 
            if (stack2.length != 0) {
                b = stack2.pop();
            }
 
            var total = a + b + carry;
 
    var temp = new Node(total % 10);
            carry = parseInt(total / 10);
 
            if (result == null) {
                result = temp;
            } else {
                temp.next = result;
                result = temp;
            }
        }
 
        if (carry != 0) {
    var temp = new Node(carry);
            temp.next = result;
            result = temp;
        }
        return result;
    }
 
    // To print a linked list
    function printList() {
        while (result != null) {
            document.write(result.data + " ");
            result = result.next;
        }
        document.write();
    }
 
    // Driver code
     
        var arr1 = [ 5, 6, 7 ];
        var arr2 = [ 1, 8 ];
 
        var size1 = 3;
        var size2 = 2;
 
        // Create first list as 5->6->7
        var i;
        for (var i = size1 - 1; i >= 0; --i)
            push(arr1[i]);
 
        // Create second list as 1->8
        for (i = size2 - 1; i >= 0; --i)
            push1(arr2[i]);
 
        result = addTwoNumbers();
 
        printList();
 
// This code contributed by umadevi9616 
</script>

Output

5 8 5

Time Complexity: O(m+n) where m and n are the sizes of given two linked lists.
Auxiliary Space: O(m+n)

Another Approach (Simple iteration with 2 pointers)

To add two numbers in a linked list, we can simply iterate the 2 linked lists and take the sum of the values in nodes along with maintaining a carry. While taking sums add the previous carry to it and add a new node to the result containing the last digit in the sum and update the carry for the next iteration.

Algorithm :

Use 2 pointers to store the head of each linked list and initialize carry as 0.
Declare a pointer to the node to store our answer.
Iterate through both the linked list and add the digits pointed by the pointers (if we have reached the end of one of the linked lists and have no further nodes, consider its value as 0 while taking the sum).
Add a new node with ((sum+carry)%10) as value to our answer and update carry as (sum + carry)/10.
Repeat steps 3 and 4 till we reach the end of both the linked lists.
After traversing both the linked lists, if carry > 0 then add this carry to our answer as a new node.

C++

#include<bits/stdc++.h>
using namespace std;
 
struct Node {
    int val;
    Node* next;
 
    Node(int value){
        val = value;
        next = NULL;
    }
};
 
void push_front(Node** head, int new_val){
    Node* new_node = new Node(new_val);
    new_node->next = *head;
    *head = new_node;
}
 
void printList(Node* head){
    Node* i = head;
    while(i){
        cout<<i->val<<" ";
        i = i->next;
    }
    cout<<"\n";
}
 
// function to add 2 numbers given as linked lists
Node* add(Node* l1, Node* l2){
    Node* ans = new Node(0);
    Node* curr = ans;
    int carry = 0;
    while(l1 || l2){
        int sum = 0;
        if(l1) sum += l1->val;
        if(l2) sum += l2->val;
        sum += carry;
 
        curr->next = new Node(sum%10);
        curr = curr->next;
 
        carry = sum/10;
 
        if(l1) l1 = l1->next;
        if(l2) l2 = l2->next;
    }
 
    if(carry){
        curr->next = new Node(carry);
    }
    ans = ans->next;
    return ans;
}
 
int main(){
    Node* l1 = NULL;
 
    push_front(&l1, 1);
    push_front(&l1, 5);
 
    Node* l2 = NULL;
 
    push_front(&l2, 3);
    push_front(&l2, 9);
    push_front(&l2, 7);
 
    // l1-> 5 1   =  15
    // l2-> 7 9 3 = 397
 
    Node* sum = add(l1,l2);
 
    printList(sum);
    // 2 1 4      = 412
}

Java

/*package whatever //do not write package name here */
 
class Node 
{
    int val;
    Node next;
    Node(int value)
    {
        val = value;
        next = null;
    }
}
class AddTwoLL 
{
    static void push_front(Node head, int new_val)
    {
        Node new_node = new Node(new_val);
        new_node.next =head;
        head = new_node;
    }
    static void printList(Node head)
    {
        Node i = head;
        while(i!=null)
        {
            System.out.print(i.val+" ");
            i = i.next;
        }
        System.out.println();
    }
    // function to reverse a linked list
    static Node reverse_it(Node head)
    {
        Node prev = null;
        Node curr = head, next;
        while(curr!=null)
        {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
    // function to convert a linked list to integer
    static int to_integer(Node head){
        int num = 0;
        Node curr = head;
        while(curr!=null)
        {
            int digit = curr.val;
            num = num*10 + digit;
            curr = curr.next;
        }
        return num;
    }
    // function to convert a number to a linked list containing digits in reverse order
    static Node to_linkedlist(int x)
    {
        Node head = new Node(0);
        if(x==0) return head;
        Node curr = head;
        while(x>0)
        {
            int d = x%10;
            x /= 10;
            curr.next = new Node(d);
            curr = curr.next;
        }
        return head.next;
    }
    // function to add 2 numbers given as linked lists
    static Node add_list(Node l1, Node l2)
    {
        // reversing the 2 linked lists
        l1 = reverse_it(l1);
        l2 = reverse_it(l2);
     
        // converting them into integers
        int num1 = to_integer(l1);
        int num2 = to_integer(l2);
         
        int sum = num1 + num2;
        // converting the sum back to
        // a linked list
        Node ans = to_linkedlist(sum);
         
        return ans;
    }
    public static void main(String [] args)
    {
        Node l1 = null;
     
        push_front(l1, 1);
        push_front(l1, 5);
     
        Node l2 = null;
     
        push_front(l2, 3);
        push_front(l2, 9);
        push_front(l2, 7);
     
        // l1-> 5 1
        // l2-> 7 9 3
     
        Node sum = add_list(l1,l2);
     
        printList(sum);
        // 2 1 4
    }
}

Python3

# Add two numbers represented by linked lists | Set 2
class Node:
    def __init__(self, value):
        self.val = value
        self.next = None
 
 
def push_front(head, new_val):
    new_node = Node(new_val)
    new_node.next = head
    head = new_node
    return head
 
 
def print_list(head):
    i = head
    while i:
        print(i.val, end=" ")
        i = i.next
    print()
 
 
def add(l1, l2):
    ans = Node(0)
    curr = ans
    carry = 0
 
    while l1 or l2:
        sum_val = 0
        if l1:
            sum_val += l1.val
            l1 = l1.next
        if l2:
            sum_val += l2.val
            l2 = l2.next
        sum_val += carry
 
        curr.next = Node(sum_val % 10)
        curr = curr.next
        carry = sum_val // 10
 
    if carry:
        curr.next = Node(carry)
 
    ans = ans.next
    return ans
 
 
# Creating linked list l1: 5 -> 1
l1 = None
l1 = push_front(l1, 1)
l1 = push_front(l1, 5)
 
# Creating linked list l2: 7 -> 9 -> 3
l2 = None
l2 = push_front(l2, 3)
l2 = push_front(l2, 9)
l2 = push_front(l2, 7)
 
# l1: 5 -> 1   =  15
# l2: 7 -> 9 -> 3 = 397
 
# Adding two linked lists
sum_list = add(l1, l2)
 
# Printing the resulting linked list
print_list(sum_list)
# Output: 2 1 4   = 412
 
# This code is contributed by uomkar369

Output

2 1 4

Time Complexity: O(n).

Space Complexity: O(max(n,m))

To avoid the use of extra space we can store the sum in one of the linked lists itself.

Related Article: Add two numbers represented by linked lists | Set 1
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