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. 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  // represents number 563
Second List: 8->4->2 //  represents number 842
Output
Resultant list: 1->4->0->5  // represents number 1405
```

We have discussed a solution here which is for linked lists where least significant digit is first node of lists and most significant digit is last node. In this problem, most significant node is first node and least significant digit is 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 sum of rightmost nodes and forward carry to left side.
3) If size is not same, then follow below steps:
….a) Calculate difference of sizes of two linked lists. Let the difference be diff
….b) Move diff nodes ahead in the bigger linked list. Now use step 2 to calculate sum of smaller list and right sub-list (of same size) of larger list. Also, store the carry of this sum.
….c) Calculate sum of the carry (calculated in previous step) with the remaining left sub-list of larger list. Nodes of this sum are added at the beginning of sum list obtained previous step.

Following is implementation of the above approach.

## C

 `// A recursive program to add two linked lists ` ` `  `#include ` `#include ` ` `  `// 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 linked list */` `void` `push(``struct` `Node** head_ref, ``int` `new_data) ` `{ ` `    ``/* allocate node */` `    ``struct` `Node* new_node = (``struct` `Node*) ``malloc``(``sizeof``(``struct` `Node)); ` ` `  `    ``/* put in the data  */` `    ``new_node->data  = new_data; ` ` `  `    ``/* link the old list off 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(``struct` `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 size of linked list */` `int` `getSize(``struct` `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 = (Node *)``malloc``(``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; ` ` `  `    ``// Assigne 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 fron of ` `    ``// the result list. e.g. 999 and 87 ` `    ``if` `(carry) ` `        ``push(result, carry); ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `    ``Node *head1 = NULL, *head2 = NULL, *result = NULL; ` ` `  `    ``int` `arr1[] = {9, 9, 9}; ` `    ``int` `arr2[] = {1, 8}; ` ` `  `    ``int` `size1 = ``sizeof``(arr1) / ``sizeof``(arr1); ` `    ``int` `size2 = ``sizeof``(arr2) / ``sizeof``(arr2); ` ` `  `    ``// 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; ` `} `

/div>

## Java

 `// Java 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 `

Output:

`1  0  1  7`

Time Complexity: O(m+n) where m and n are the sizes of given two linked lists.

Related Article : Add two numbers represented by linked lists | Set 1