# Multiply two numbers represented by Linked Lists ### Introduction

The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Proper understanding of concepts based on Linked Lists can give you an edge in a coding interview.

### Problem Statement

In this problem, we are given two linked lists, L1 and L2. Both L1 and L2 are numbers represented as linked lists. We are required to multiply these Lists and produce the output.

### Problem Statement Understanding

Let’s try to understand the problem with help of examples by referring the best online learning sites for programming.

Suppose the given linked lists are:  Now, according to the problem statement, these linked lists L1 and L2 are actually two numbers which are represented using linked lists.
For example, we can represent

• The number 23 as 2→3 using linked list.
• The number 234 as 2→3→4 using linked list.
• More generically speaking, we can represent a number abcd as a→b→c→d using linked list.

And our output is the multiplication of these two numbers L1 and L2, represented as a linked list.

• As L1 = 5→6→1, so the number is 561.
• And as L2 = 4→2, the number is 42.

Output = 561*42 = 23562

Let’s say if:  • Corresponding numbers for L1 and L2 will be 2345 and 345.

Output = 2345*345 = 809025

#### Some more examples:

Input: L1 = 3→2→1, L2 = 2→1
Output: 6741

Input: L1 = 3→2, L2 = 1→7
Output: 544

Input: L1 = 9→8→7, L2 = 1→2→3
Output: 121401

Now, I hope that from the above examples, you got the idea about what the problem is. So now let’s move towards finding some approach to solve this problem.

### Approach

Firstly, traverse through both the lists and produce the numbers required to be multiplied, and then return the multiplied values of the two numbers.

##### Algorithm for producing the number from linked list representation:
• Initialize a variable num to zero.
• Begin traversing through the linked list.
• Add the data of the first node to this variable num.
• Second node onwards,
1) multiply the variable num by 10
2) and take the modulus of this value by 109+7 as well,
3) and then add the data of the node to the variable num.
• Repeat the previous step until we arrive at the last node of the list.

First, using the above-discussed algorithm for producing the number from linked list representation, we will produce the required numbers from the given linked lists L1 and L2.

After the numbers have been produced, we will multiply them to get the output.

### Dry Run  ### Code Implementation

```#include
#include

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

// Function to create a new node
// with given data
struct Node *newNode(int data)
{
struct Node *new_node = (struct Node *) malloc(sizeof(struct Node));
new_node->data = data;
new_node->next = NULL;
return new_node;
}

// Function to insert a node at the
// beginning of the Linked List
void push(struct Node** head_ref, int new_data)
{
// allocate node
struct Node* new_node = newNode(new_data);

// link the old list off the new node

// move the head to point to the new node
}

// Multiply contents of two linked lists
long long multiplyTwoLists (struct Node* first,struct Node* second)
{
long long N= 1000000007;
long long num1 = 0, num2 = 0;
while (first || second){

if(first){
num1 = ((num1)*10)%N + first->data;
first = first->next;
}

if(second)
{
num2 = ((num2)*10)%N + second->data;
second = second->next;
}

}
return ((num1%N)*(num2%N))%N;
}

// A utility function to print a linked list
void printList(struct Node *node)
{
while(node != NULL)
{
printf("%d",node->data);
if(node->next)
printf("->");
node = node->next;
}
printf("\n");
}

// Driver program to test above function
int main()
{
struct Node* first = NULL;
struct Node* second = NULL;

// create first list 9->4->6
push(&first, 6);
push(&first, 4);
push(&first, 9);
printf("First List is: ");
printList(first);

// create second list 8->4
push(&second, 4);
push(&second, 8);
printf("Second List is: ");
printList(second);

// Multiply the two lists and see result
printf("Result is: ");
long long ans = multiplyTwoLists(first, second);
printf("%lld",ans);

return 0;
}
```
```#include
#include
using namespace std;

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

// Function for creating a new node with given value
struct Node *create_node(int val)
{
struct Node *node_new = (struct Node *)
malloc(sizeof(struct Node));

// putting the data into the new node.
node_new->val = val;
// initially new node points to NULL
node_new->next = NULL;

return node_new;
}

// Function for inserting a node
// at the start of the Linked List
void push(struct Node** head_ref, int new_val)
{
// assigning the node
struct Node* node_new = create_node(new_val);

// Now, link old list off the new node

// moving the head inorder to point to the new node
}

// to multiply data present in the two linked lists
long long multiplication (Node* list_one, Node* list_two)
{
long long n= 1000000007;
long long first_num = 0, second_num = 0;
while (list_one || list_two){

if(list_one){
first_num = ((first_num)*10)%n + list_one->val;
list_one = list_one->next;
}

if(list_two)
{
second_num = ((second_num)*10)%n + list_two->val;
list_two = list_two->next;
}

}
return ((first_num%n)*(second_num%n))%n;
}

// For printing the linked list
void display_list(struct Node *node)
{
while(node != NULL)
{
cout<val;
if(node->next)
cout<<"->";
node = node->next;
}
}

// Driver function
int main()
{
struct Node* one = NULL;
struct Node* two = NULL;

// creating list one 5->6->1
push(&one, 1);
push(&one, 6);
push(&one, 5);
printf("List one is: ");
display_list(one);

cout<<”\n”;

// creating list two 4->2
push(&two, 2);
push(&two, 4);
printf("List two is: ");
display_list(two);

cout<<”\n”;

// Results after multiplying the two numbers.

```
```class Multiply
{
static class Node
{
int data;
Node next;

Node(int data){
this.data = data;
next = null;
}
}
// Multiply contents of two linked lists
static long multiplyTwoLists(Node first, Node second)
{
long N = 1000000007;
long num1 = 0, num2 = 0;

while (first != null || second != null){

if(first != null){
num1 = ((num1)*10)%N + first.data;
first = first.next;
}

if(second != null)
{
num2 = ((num2)*10)%N + second.data;
second = second.next;
}

}
return ((num1%N)*(num2%N))%N;
}
static void printList(Node node)
{
while(node != null)
{
System.out.print(node.data);
if(node.next != null)
System.out.print("->");
node = node.next;
}
System.out.println();
}

// Driver program to test above function
public static void main(String args[])
{
// create first list 9->4->6
Node first = new Node(9);
first.next = new Node(4);
first.next.next = new Node(6);
System.out.print("First List is: ");
printList(first);

// create second list 8->4
Node second = new Node(8);
second.next = new Node(4);
System.out.print("Second List is: ");
printList(second);

// Multiply the two lists and see result
System.out.print("Result is: ");
System.out.println(multiplyTwoLists(first, second));
}
}
```
```
class Node:

def __init__(self, data):

self.data = data
self.next = None

def __init__(self):

# Function to insert a node at the
# beginning of the Linked List
def push(self, new_data):

# Create a new Node
new_node = Node(new_data)

# Make next of the new Node as head

# Move the head to point to new Node

# Function to print the Linked List
def printList(self):

while (ptr != None):
print(ptr.data, end = '')
if ptr.next != None:
print('->', end = '')

ptr = ptr.next

print()

# Multiply contents of two Linked Lists
def multiplyTwoLists(first, second):

num1 = 0
num2 = 0

while first_ptr != None or second_ptr != None:
if first_ptr != None:
num1 = (num1 * 10) + first_ptr.data
first_ptr = first_ptr.next

if second_ptr != None:
num2 = (num2 * 10) + second_ptr.data
second_ptr = second_ptr.next

return num1 * num2

# Driver code
if __name__=='__main__':

first.push(1)
first.push(6)
first.push(5)

print("First list is: ", end = '')
first.printList()

second.push(2)
second.push(4)

print("Second List is: ", end = '')
second.printList()

result = multiplyTwoLists(first, second)
print("Result is: ", result)
```

### Output

List one is: 5→6→1
List two is: 4→2