Given two numbers represented by two linked lists, the task is to 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.
Step-by-step approach:
- Calculate sizes of given two linked lists.
- 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.
-
If size is not same, then follow below steps:
- Calculate difference of sizes of two linked lists. Let the difference be diff
- 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.
- 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.
// 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 |
// 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 the linked list */ void push(node** head_ref, int new_data)
{ /* allocate node */
node* new_node = (node*) malloc ( sizeof (node));
/* put in the data */
new_node->data = new_data;
/* link the old list to 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 a linked list */ void printList(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 the size of a linked list */ int getSize(node* node)
{ int size = 0;
while (node != NULL) {
node = node->next;
size++;
}
return size;
} // Adds two linked lists of the same // size represented by head1 // and head2 and returns the 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 the same
// size, check any of the two
// head pointers
if (head1 == NULL)
return NULL;
int sum;
// Allocate memory for the sum
// node of the current two nodes
node* result = (node*) malloc ( sizeof (node));
// Recursively add the remaining nodes
// and get the carry
result->next
= addSameSize(head1->next, head2->next, carry);
// add digits of the current nodes
// and propagated carry
sum = head1->data + head2->data + *carry;
*carry = sum / 10;
sum = sum % 10;
// Assign the sum to the current
// node of the resultant list
result->data = sum;
return result;
} // This function is called after // the smaller list is added // to the bigger list's sublist // of the same size. Once the // right sublist is added, the // carry must be added to the left // side of the larger list to get // the final result. void addCarryToRemaining(node* head1, node* cur, int * carry,
node** result)
{ int sum;
// If a different 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 to 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 lists of the same size
if (size1 == size2)
*result = addSameSize(head1, head2, &carry);
else {
int diff = abs (size1 - size2);
// First list should always be
// larger than the second list.
// If not, swap pointers
if (size1 < size2)
swapPointer(&head1, &head2);
// move the diff. number of nodes in the first list
for (cur = head1; diff--; cur = cur->next)
;
// get the addition of the same size lists
*result = addSameSize(cur, head2, &carry);
// get the addition of the 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;
node* head2 = NULL;
node* 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 the first list as 9->9->9
int i;
for (i = size1 - 1; i >= 0; --i)
push(&head1, arr1[i]);
// Create the second list as 1->8
for (i = size2 - 1; i >= 0; --i)
push(&head2, arr2[i]);
addList(head1, head2, &result);
printf ( "Resultant List: " );
printList(result);
return 0;
} |
// 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 |
# 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 |
// 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 |
<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> |
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++ 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 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 |
# 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# 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 |
<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> |
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.
Step-by-step approach:
- 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.
Below is the implementation of the above approach:
#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
} |
/*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
}
} |
# 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 |
using System;
class Node
{ public int val;
public Node next;
public Node( int value)
{
val = value;
next = null ;
}
} class AddTwoLL
{ static void push_front( ref 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 )
{
Console.Write(i.val + " " );
i = i.next;
}
Console.WriteLine();
}
// Function to reverse a linked list
static Node reverse_it(Node head)
{
Node prev = null ;
Node curr = head;
Node 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( ref l1, 1);
push_front( ref l1, 5);
Node l2 = null ;
push_front( ref l2, 3);
push_front( ref l2, 9);
push_front( ref l2, 7);
// l1-> 5 1
// l2-> 7 9 3
Node sum = add_list(l1, l2);
printList(sum);
// 2 1 4
}
} |
class Node { constructor(value) {
this .val = value;
this .next = null ;
}
} function pushFront(head, newVal) {
const newNode = new Node(newVal);
newNode.next = head;
return newNode;
} function printList(head) {
let current = head;
while (current) {
process.stdout.write(current.val + " " );
current = current.next;
}
console.log();
} // Function to add two numbers given as linked lists function add(l1, l2) {
const ans = new Node(0);
let curr = ans;
let carry = 0;
while (l1 || l2) {
let sum = 0;
if (l1) sum += l1.val;
if (l2) sum += l2.val;
sum += carry;
curr.next = new Node(sum % 10);
curr = curr.next;
carry = Math.floor(sum / 10);
if (l1) l1 = l1.next;
if (l2) l2 = l2.next;
}
if (carry) {
curr.next = new Node(carry);
}
return ans.next;
} // Main function function main() {
let l1 = null ;
l1 = pushFront(l1, 1);
l1 = pushFront(l1, 5);
let l2 = null ;
l2 = pushFront(l2, 3);
l2 = pushFront(l2, 9);
l2 = pushFront(l2, 7);
// l1-> 5 1 = 15
// l2-> 7 9 3 = 397
const sum = add(l1, l2);
printList(sum);
// 2 1 4 = 412
} main(); |
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