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Javascript 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. 

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 code
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 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.

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 extra node at the start of the resultant list to accommodate this carry.

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 off 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 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
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 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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