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

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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 C implementation of the above approach.

// A recursive program to add two linked lists #include <stdio.h> #include <stdlib.h> // 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[0]); int size2 = sizeof(arr2) / sizeof(arr2[0]); // 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; }

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

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