Add two numbers represented by linked lists | Set 1

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

Example 1

Input:
  First List: 5->6->3  // represents number 365
  Second List: 8->4->2 //  represents number 248
Output
  Resultant list: 3->1->6  // represents number 613

Example 2

Input:
  First List: 7->5->9->4->6  // represents number 64957
  Second List: 8->4 //  represents number 48
Output
  Resultant list: 5->0->0->5->6  // represents number 65005

Solution
Traverse both lists. One by one pick nodes of both lists and add the values. If sum is more than 10 then make carry as 1 and reduce sum. If one list has more elements than the other then consider remaining values of this list as 0. Following is the implementation of this approach.

C

#include<stdio.h>
#include<stdlib.h>

/* Linked list node */
struct node
{
    int data;
    struct node* next;
};

/* Function to create a new node with given data */
struct node *newNode(int data)
{
    struct node *new_node = (struct node *) malloc(sizeof(struct node));
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}

/* Function to insert a node at the beginning of the Doubly Linked List */
void push(struct node** head_ref, int new_data)
{
    /* allocate node */
    struct node* new_node = newNode(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;
}

/* Adds contents of two linked lists and return the head node of resultant list */
struct node* addTwoLists (struct node* first, struct node* second)
{
    struct node* res = NULL; // res is head node of the resultant list
    struct node *temp, *prev = NULL;
    int carry = 0, sum;

    while (first != NULL || second != NULL) //while both lists exist
    {
        // Calculate value of next digit in resultant list.
        // The next digit is sum of following things
        // (i)  Carry
        // (ii) Next digit of first list (if there is a next digit)
        // (ii) Next digit of second list (if there is a next digit)
        sum = carry + (first? first->data: 0) + (second? second->data: 0);

        // update carry for next calulation
        carry = (sum >= 10)? 1 : 0;

        // update sum if it is greater than 10
        sum = sum % 10;

        // Create a new node with sum as data
        temp = newNode(sum);

        // if this is the first node then set it as head of the resultant list
        if(res == NULL)
            res = temp;
        else // If this is not the first node then connect it to the rest.
            prev->next = temp;

        // Set prev for next insertion
        prev  = temp;

        // Move first and second pointers to next nodes
        if (first) first = first->next;
        if (second) second = second->next;
    }

    if (carry > 0)
      temp->next = newNode(carry);

    // return head of the resultant list
    return res;
}

// A utility function to print a linked list
void printList(struct node *node)
{
    while(node != NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
    printf("\n");
}

/* Driver program to test above function */
int main(void)
{
    struct node* res = NULL;
    struct node* first = NULL;
    struct node* second = NULL;

    // create first list 7->5->9->4->6
    push(&first, 6);
    push(&first, 4);
    push(&first, 9);
    push(&first, 5);
    push(&first, 7);
    printf("First List is ");
    printList(first);

    // create second list 8->4
    push(&second, 4);
    push(&second, 8);
    printf("Second List is ");
    printList(second);

    // Add the two lists and see result
    res = addTwoLists(first, second);
    printf("Resultant list is ");
    printList(res);

   return 0;
}

Java

// Java program to delete a given node in linked list under given constraints

class LinkedList {

    static Node head1, head2;

    static class Node {

        int data;
        Node next;

        Node(int d) {
            data = d;
            next = null;
        }
    }

    /* Adds contents of two linked lists and return the head node of resultant list */
    Node addTwoLists(Node first, Node second) {
        Node res = null; // res is head node of the resultant list
        Node prev = null;
        Node temp = null;
        int carry = 0, sum;

        while (first != null || second != null) //while both lists exist
        {
            // Calculate value of next digit in resultant list.
            // The next digit is sum of following things
            // (i)  Carry
            // (ii) Next digit of first list (if there is a next digit)
            // (ii) Next digit of second list (if there is a next digit)
            sum = carry + (first != null ? first.data : 0)
                    + (second != null ? second.data : 0);

            // update carry for next calulation
            carry = (sum >= 10) ? 1 : 0;

            // update sum if it is greater than 10
            sum = sum % 10;

            // Create a new node with sum as data
            temp = new Node(sum);

            // if this is the first node then set it as head of
            // the resultant list
            if (res == null) {
                res = temp;
            } else // If this is not the first node then connect it to the rest.
            {
                prev.next = temp;
            }

            // Set prev for next insertion
            prev = temp;

            // Move first and second pointers to next nodes
            if (first != null) {
                first = first.next;
            }
            if (second != null) {
                second = second.next;
            }
        }

        if (carry > 0) {
            temp.next = new Node(carry);
        }

        // return head of the resultant list
        return res;
    }
    /* Utility function to print a linked list */

    void printList(Node head) {
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
        System.out.println("");
    }

    public static void main(String[] args) {
        LinkedList list = new LinkedList();

        // creating first list
        list.head1 = new Node(7);
        list.head1.next = new Node(5);
        list.head1.next.next = new Node(9);
        list.head1.next.next.next = new Node(4);
        list.head1.next.next.next.next = new Node(6);
        System.out.print("First List is ");
        list.printList(head1);

        // creating seconnd list
        list.head2 = new Node(8);
        list.head2.next = new Node(4);
        System.out.print("Second List is ");
        list.printList(head2);

        // add the two lists and see the result
        Node rs = list.addTwoLists(head1, head2);
        System.out.print("Resultant List is ");
        list.printList(rs);
    }
}

// this code has been contributed by Mayank Jaiswal

Python


# Python program to add two numbers represented by linked list

# Node class
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Add contents of two linked lists and return the head
    # node of resultant list
    def addTwoLists(self, first, second):
        prev = None
        temp = None
        carry = 0 

        # While both list exists
        while(first is not None or second is not None):

            # Calculate the value of next digit in
            # resultant list
            # The next digit is sum of following things
            # (i) Carry
            # (ii) Next digit of first list (if ther is a
            # next digit)
            # (iii) Next digit of second list ( if there
            # is a next digit)
            fdata = 0 if first is None else first.data
            sdata = 0 if second is None else second.data
            Sum = carry + fdata + sdata

            # update carry for next calculation
            carry = 1 if Sum >= 10 else 0

            # update sum if it is greater than 10
            Sum = Sum if Sum < 10 else Sum % 10

            # Create a new node with sum as data
            temp = Node(Sum)

            # if this is the first node then set it as head
            # of resultant list
            if self.head is None:
                self.head = temp
            else :
                prev.next = temp 

            # Set prev for next insertion
            prev = temp

            # Move first and second pointers to next nodes
            if first is not None:
                first = first.next
            if second is not None:
                second = second.next

        if carry > 0:
            temp.next = Node(carry)

    # Utility function to print the linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print temp.data,
            temp = temp.next

# Driver program to test above function
first = LinkedList()
second = LinkedList()

# Create first list
first.push(6)
first.push(4)
first.push(9)
first.push(5)
first.push(7)
print "First List is ",
first.printList()

# Create second list
second.push(4)
second.push(8)
print "\nSecond List is ",
second.printList()

# Add the two lists and see result
res = LinkedList()
res.addTwoLists(first.head, second.head)
print "\nResultant list is ",
res.printList()

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)


Output:

  First List is 7 5 9 4 6
  Second List is 8 4
  Resultant list is 5 0 0 5 6

Time Complexity: O(m + n) where m and n are number of nodes in first and second lists respectively.

Related Article : Add two numbers represented by linked lists | Set 2

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

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