Add two numbers represented by Linked List without any extra space

Given two numbers represented by two linked lists, write a function that returns sum list. The sum list is linked list representation of addition of two input numbers. Expected Space Complexity O(1).

Examples:

Input:
L1 = 5 -> 6 -> 3 -> NULL 
L2 = 8 -> 4 -> 2 -> NULL 
Output: 1 -> 4 -> 0 -> 5 -> NULL

Input:
L1 = 1 -> 0 -> 0 -> NULL 
L2 = 9 -> 1 -> NULL 
Output: 1 -> 9 -> 1 -> NULL 

Approach: We have discussed a solution here where we used recursion to reach to the least significant number in the lists, but due to the involvement of the stack, the space complexity of the solution becomes O(N)
Here the target is to do the sum inplace, and return the modified sum list.

The idea is to first reverse both the linked list, so new head of the list points to least significant number and we can start adding as described here and instead of creating a new list, we modify the existing one and return the head of modified list.

Following are the steps: 

1. Reverse List L1.
2. Reverse List L2.
3. Add the nodes of both the lists iteratively.
4. Reverse the resultant list and return its head.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

class LinkedList;

// Node class for the linked list
class Node {
    int data;
    Node* next;

    friend LinkedList;

public:
    Node();
    Node(int x);
};

Node::Node()
{
    data = 0;
    next = NULL;
}

// Function to initialise
// a node with value x
Node::Node(int x)
{
    data = x;
    next = NULL;
}

// Linkedlist class with helper functions
class LinkedList {
public:
    Node* head;

    LinkedList();

    void insert(int x);
    void reverse();
    void traverse();

    void sum(LinkedList*);
};

LinkedList::LinkedList()
{
    head = NULL;
}

// Function to insert a node at
// the head of the list
void LinkedList::insert(int x)
{
    Node* node = new Node();
    node->data = x;

    if (head == NULL)
        head = node;
    else {
        node->next = head;
        head = node;
    }
}

// Function to reverse the linked list
void LinkedList::reverse()
{
    Node *prev = NULL, *curr = head;

    while (curr) {
        Node* temp = curr->next;
        curr->next = prev;
        prev = curr;
        curr = temp;
    }
    head = prev;
}

// Function to traverse and print the list
void LinkedList::traverse()
{

    Node* temp = head;

    while (temp) {
        cout << temp->data << " -> ";
        temp = temp->next;
    }
    cout << "NULL";
}

// Function to add two numbers
// represented as linked lists
void LinkedList::sum(LinkedList* l2)
{
    reverse();
    l2->reverse();

    Node *start1 = head, *start2 = l2->head;

    Node* prev = NULL;
    int carry = 0;

    // While both lists exist
    while (start1 && start2) {

        // Current sum
        int temp = start1->data + start2->data + carry;

        // Handle carry
        start1->data = temp % 10;
        carry = temp / 10;
        prev = start1;

        // Get to next nodes
        start1 = start1->next;
        start2 = start2->next;
    }

    // If there are remaining digits
    // in any one of the lists
    if (start1 || start2) {

        if (start2)
            prev->next = start2;

        start1 = prev->next;

        // While first list has digits remaining
        while (start1) {
            int temp = start1->data + carry;
            start1->data = temp % 10;
            carry = temp / 10;
            prev = start1;
            start1 = start1->next;
        }
    }

    // If a new node needs to be
    // created due to carry
    if (carry > 0) {
        prev->next = new Node(carry);
    }

    // Reverse the resultant list
    reverse();
}

// Driver code
int main()
{

    // Create first list
    LinkedList* l1 = new LinkedList();
    l1->insert(3);
    l1->insert(6);
    l1->insert(5);

    // Create second list
    LinkedList* l2 = new LinkedList();
    l2->insert(2);
    l2->insert(4);
    l2->insert(8);

    // Add the lists
    l1->sum(l2);

    // Print the resultant list
    l1->traverse();

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
class Node
{
    int data;
    Node next;

    // constructor
    Node(int d)
    {
        data = d;
        next = null;
    }
}// Node closes

class LinkedList
{
    Node head;

    // Helper function to traverse
    void traverse(Node head)
    {
        while(head != null)
        {
            System.out.print(head.data + "->");
            head = head.next;
        }
    }

    // Helper function to insert data in linked list
    void insert(int x)
    {
        Node temp = new Node(x);
        if(head == null) head = temp;
        else
        {
            temp.next = head;
            head = temp;
        }
    }

    // Helper function to reverse the list
    public static Node reverse(Node head)
    {
        if(head == null || head.next == null) return head;

        Node prev = null;
        Node curr = head;
        while(curr != null)
        {
            Node temp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = temp;
        }
        head = prev;
        return head;
    }

    // Function to add two lists
    public static Node sum(Node l1, Node l2)
    {
        if(l2 == null) return l1;
        if(l1 == null) return l2;

        // reverse l1 list
        l1 = reverse(l1);

        // reverse l2 list
        l2 = reverse(l2);

        // storing head whose reverse is to be returned
        // This is where which will be final node
        Node head = l1;
        Node prev = null;
        int c = 0,sum;
        while(l1 != null && l2 != null)
        {
            sum = c + l1.data + l2.data;
            l1.data = sum % 10;
            c = sum / 10;
            prev = l1;
            l1 = l1.next;
            l2 = l2.next;
        }

        if(l1 != null||l2 != null)
        {
            if(l2 != null) prev.next = l2;
            l1 = prev.next;
            while(l1 != null)
            {
                sum = c + l1.data;
                l1.data = sum % 10;
                c = sum / 10;
                prev = l1;
                l1 = l1.next;
            }
        }
        if(c > 0) prev.next = new Node(c);
        return reverse(head);
    }

    // Driver Code
    public static void main(String[] args)
    {
        LinkedList l1 = new LinkedList();
        l1.insert(3);
        l1.insert(6);
        l1.insert(5);
        LinkedList l2 = new LinkedList();
        l2.insert(2);
        l2.insert(4);
        l2.insert(8);
        LinkedList l3 = new LinkedList();
        Node head = sum(l1.head, l2.head);
        l3.traverse(head);
        System.out.print("Null");
    }
}

// This code is contributed
// by Devarshi Singh

Python3

# Python3 implementation of the approach

# Linked List Node
class Node:

    def __init__(self, data):

        self.data = data
        self.next = None

# Handle list operations
class LinkedList:

    def __init__(self):

        self.head = None

    # Method to traverse list and
    # return it in a format
    def traverse(self):

        linkedListStr = ""
        temp = self.head

        while temp:
            linkedListStr += str(temp.data) + " -> "
            temp = temp.next

        return linkedListStr + "NULL"

    # Method to insert data in linked list
    def insert(self, data):

        newNode = Node(data)

        if self.head is None:
            self.head = newNode
        else:
            newNode.next = self.head
            self.head = newNode

# Helper function to reverse the list
def reverse(Head):

    if (Head is None and
        Head.next is None):
        return Head

    prev = None
    curr = Head

    while curr:
        temp = curr.next
        curr.next = prev
        prev = curr
        curr = temp

    Head = prev
    return Head

# Function to add two lists
def listSum(l1, l2):

    if l1 is None:
        return l1
    if l2 is None:
        return l2

    # Reverse first list
    l1 = reverse(l1)

    # Reverse second list
    l2 = reverse(l2)

    # Storing head whose reverse
    # is to be returned This is
    # where which will be final node
    head = l1
    prev = None
    c = 0
    sum = 0

    while l1 is not None and l2 is not None:
        sum = c + l1.data + l2.data
        l1.data = sum % 10
        c = int(sum / 10)
        prev = l1
        l1 = l1.next
        l2 = l2.next

    if l1 is not None or l2 is not None:
        if l2 is not None:
            prev.next = l2
        l1 = prev.next

        while l1 is not None:
            sum = c + l1.data
            l1.data = sum % 10
            c = int(sum / 10)
            prev = l1
            l1 = l1.next

    if c > 0:
        prev.next = Node(c)

    return reverse(head)

# Driver code
linkedList1 = LinkedList()
linkedList1.insert(3)
linkedList1.insert(6)
linkedList1.insert(5)

linkedList2 = LinkedList()
linkedList2.insert(2)
linkedList2.insert(4)
linkedList2.insert(8)

linkedList3 = LinkedList()
linkedList3.head = listSum(linkedList1.head,
                           linkedList2.head)

print(linkedList3.traverse())

# This code is contributed by Debidutta Rath

C#

// C# implementation of the above approach
using System;

public class Node
{
    public int data;
    public Node next;

    // constructor
    public Node(int d)
    {
        data = d;
        next = null;
    }
}

// Node closes
public class LinkedList
{
    Node head;

    // Helper function to traverse
    void traverse(Node head)
    {
        while(head != null)
        {
            Console.Write(head.data + "->");
            head = head.next;
        }
    }

    // Helper function to insert data in linked list
    void insert(int x)
    {
        Node temp = new Node(x);
        if(head == null) head = temp;
        else
        {
            temp.next = head;
            head = temp;
        }
    }

    // Helper function to reverse the list
    public static Node reverse(Node head)
    {
        if(head == null ||
           head.next == null) return head;

        Node prev = null;
        Node curr = head;
        while(curr != null)
        {
            Node temp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = temp;
        }
        head = prev;
        return head;
    }

    // Function to add two lists
    public static Node sum(Node l1, Node l2)
    {
        if(l2 == null) return l1;
        if(l1 == null) return l2;

        // reverse l1 list
        l1 = reverse(l1);

        // reverse l2 list
        l2 = reverse(l2);

        // storing head whose reverse is
        // to be returned. This is where
        // which will be final node
        Node head = l1;
        Node prev = null;
        int c = 0,sum;
        while(l1 != null && l2 != null)
        {
            sum = c + l1.data + l2.data;
            l1.data = sum % 10;
            c = sum / 10;
            prev = l1;
            l1 = l1.next;
            l2 = l2.next;
        }

        if(l1 != null||l2 != null)
        {
            if(l2 != null) prev.next = l2;
            l1 = prev.next;
            while(l1 != null)
            {
                sum = c + l1.data;
                l1.data = sum % 10;
                c = sum / 10;
                prev = l1;
                l1 = l1.next;
            }
        }
        if(c > 0) prev.next = new Node(c);
        return reverse(head);
    }

    // Driver Code
    public static void Main(String[] args)
    {
        LinkedList l1 = new LinkedList();
        l1.insert(3);
        l1.insert(6);
        l1.insert(5);
        LinkedList l2 = new LinkedList();
        l2.insert(2);
        l2.insert(4);
        l2.insert(8);
        LinkedList l3 = new LinkedList();
        Node head = sum(l1.head, l2.head);
        l3.traverse(head);
        Console.Write("Null");
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation of the approach

class Node
{
    constructor(val)
    {
        this.data = val;
        this.next = null;
    }
}

// Linkedlist class with helper functions
class LinkedList {

    constructor()
    {
        this.head = null;
    }

// Function to insert a node at
// the head of the list
insert(x)
{
    var node = new Node();
    node.data = x;

    if (this.head == null)
        this.head = node;
    else {
        node.next = this.head;
        this.head = node;
    }
}

// Function to reverse the linked list
reverse()
{
    var prev = null, curr = this.head;

    while (curr) {
        var temp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = temp;
    }
    this.head = prev;
}

// Function to traverse and print the list
traverse()
{

    var temp = this.head;

    while (temp) {
        document.write( temp.data + " -> ");
        temp = temp.next;
    }
    document.write( "null");
}

// Function to add two numbers
// represented as linked lists
sum(l2)
{
    this.reverse();
    l2.reverse();

    var start1 = this.head, start2 = l2.head;

    var prev = null;
    var carry = 0;

    // While both lists exist
    while (start1 && start2) {

        // Current sum
        var temp = start1.data + start2.data + carry;

        // Handle carry
        start1.data = temp % 10;
        carry = parseInt(temp / 10);
        prev = start1;

        // Get to next nodes
        start1 = start1.next;
        start2 = start2.next;
    }

    // If there are remaining digits
    // in any one of the lists
    if (start1 || start2) {

        if (start2)
            prev.next = start2;

        start1 = prev.next;

        // While first list has digits remaining
        while (start1) {
            var temp = start1.data + carry;
            start1.data = temp % 10;
            carry = parseInt(temp / 10);
            prev = start1;
            start1 = start1.next;
        }
    }

    // If a new node needs to be
    // created due to carry
    if (carry > 0) {
        prev.next = new Node(carry);
    }

    // Reverse the resultant list
    this.reverse();
}
};

// Driver code
// Create first list
var l1 = new LinkedList();
l1.insert(3);
l1.insert(6);
l1.insert(5);

// Create second list
var l2 = new LinkedList();
l2.insert(2);
l2.insert(4);
l2.insert(8);

// Add the lists
l1.sum(l2);

// Print the resultant list
l1.traverse();

// This code is contributed by itsok.
</script>

Output:

1 -> 4 -> 0 -> 5 -> NULL

Time Complexity: O(max(m, n)) where m and n are number of nodes in list l1 and list l2 respectively. 
Space Complexity: O(1)