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Java Program For Adding Two Numbers Represented By Linked Lists- Set 2

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  • Last Updated : 18 Jun, 2022

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

We have discussed a solution here which is for linked lists where a least significant digit is the first node of lists and the most significant digit is the last node. In this problem, the most significant node 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 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: 
    • 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 image is the implementation of the above approach.

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 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 Rishabh Mahrsee

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(n+m) due to recursive stack space

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 the stack become empty.
  • in every iteration, we keep the track of the carry.
  • In the end, if carry>0, that means we need an extra node at the start of the resultant list to accommodate this carry.

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 off 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 off 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

Output:

5 8 5

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

Related Article: Add two numbers represented by linked lists | Set 1 Please refer complete article on Add two numbers represented by linked lists | Set 2 for more details!


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