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

• 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 stack1 =``                   ``new` `Stack<>();``    ``Stack 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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