Subtract Two Numbers represented as Linked Lists

3.8

Given two linked lists that represent two large positive numbers. Subtract the smaller number from larger one and return the difference as a linked list. Note that the input lists may be in any order, but we always need to subtract smaller from larger one.

It may be assumed that there are no extra leading zeros in input lists.

Examples

Input  : l1 = 1 -> 0 -> 0 -> NULL,  l2 = 1 -> NULL
Output : 0->9->9->NULL

Input  : l1 = 1 -> 0 -> 0 -> NULL,  l2 = 1 -> NULL
Output : 0->9->9->NULL

Input  : l1 = 7-> 8 -> 6 -> NULL,  l2 = 7 -> 8 -> 9 NULL
Output : 3->NULL

Following are the steps.
1) Calculate sizes of given two linked lists.
2) If sizes not are same, then append zeros in smaller linked list.
3) If size are same, then follow below steps:
….a) Find the smaller valued linked list.
….b) One by one subtract nodes of smaller sized linked list from larger size. Keep track of borrow while subtracting.

Following is the implementation of the above approach.

C++

// C++ program to subtract smaller valued list from
// larger valued list and return result as a list.
#include<bits/stdc++.h>
using namespace std;

// A linked List Node
struct Node
{
    int data;
    struct Node* next;
};

// A utility which creates Node.
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}

/* A utility function to get length of linked list */
int getLength(Node *Node)
{
    int size = 0;
    while (Node != NULL)
    {
        Node = Node->next;
        size++;
    }
    return size;
}

/* A Utility that padds zeros in front of the
   Node, with the given diff */
Node* paddZeros(Node* sNode, int diff)
{
    if (sNode == NULL)
        return NULL;

    Node* zHead = newNode(0);
    diff--;
    Node* temp = zHead;
    while (diff--)
    {
        temp->next = newNode(0);
        temp = temp->next;
    }
    temp->next = sNode;
    return zHead;
}

/* Subtract LinkedList Helper is a recursive function,
   move till the last Node,  and subtract the digits and
   create the Node and return the Node. If d1 < d2, we
   borrow the number from previous digit. */
Node* subtractLinkedListHelper(Node* l1, Node* l2, bool& borrow)
{
    if (l1 == NULL && l2 == NULL && borrow == 0)
        return NULL;

    Node* previous = subtractLinkedListHelper(l1 ? l1->next : NULL,
                                    l2 ? l2->next : NULL, borrow);

    int d1 = l1->data;
    int d2 = l2->data;
    int sub = 0;

    /* if you have given the value value to next digit then
       reduce the d1 by 1 */
    if (borrow)
    {
        d1--;
        borrow = false;
    }

    /* If d1 < d2 , then borrow the number from previous digit.
       Add 10 to d1 and set borrow = true; */
    if (d1 < d2)
    {
        borrow = true;
        d1 = d1 + 10;
    }

    /* subtract the digits */
    sub = d1 - d2;

    /* Create a Node with sub value */
    Node* current = newNode(sub);

    /* Set the Next pointer as Previous */
    current->next = previous;

    return current;
}

/* This API subtracts two linked lists and returns the
   linked list which shall  have the subtracted result. */
Node* subtractLinkedList(Node* l1, Node* l2)
{
    // Base Case.
    if (l1 == NULL &&  l2 == NULL)
        return NULL;

    // In either of the case, get the lengths of both
    // Linked list.
    int len1 = getLength(l1);
    int len2 = getLength(l2);

    Node *lNode = NULL, *sNode = NULL;

    Node* temp1 = l1;
    Node* temp2 = l2;

    // If lengths differ, calculate the smaller Node
    // and padd zeros for smaller Node and ensure both
    // larger Node and smaller Node has equal length.
    if (len1 != len2)
    {
        lNode = len1 > len2 ? l1 : l2;
        sNode = len1 > len2 ? l2 : l1;
        sNode = paddZeros(sNode, abs(len1 - len2));
    }

    else
    {
        // If both list lengths are equal, then calculate
        // the larger and smaller list. If 5-6-7 & 5-6-8
        // are linked list, then walk through linked list
        // at last Node as 7 < 8, larger Node is 5-6-8
        // and smaller Node is 5-6-7.
        while (l1 && l2)
        {
            if (l1->data != l2->data)
            {
                lNode = l1->data > l2->data ? temp1 : temp2;
                sNode = l1->data > l2->data ? temp2 : temp1;
                break;
            }
            l1 = l1->next;
            l2 = l2->next;
        }
    }

    // After calculating larger and smaller Node, call
    // subtractLinkedListHelper which returns the subtracted
    // linked list.
    bool borrow = false;
    return subtractLinkedListHelper(lNode, sNode, borrow);
}

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

// Driver program to test above functions
int main()
{
    Node* head1 = newNode(1);
    head1->next = newNode(0);
    head1->next->next = newNode(0);

    Node* head2 = newNode(1);

    Node* result = subtractLinkedList(head1, head2);

    printList(result);

    return 0;
}

Java

// Java program to subtract smaller valued 
// list from larger valued list and return 
// result as a list.
import java.util.*;
import java.lang.*;
import java.io.*;

class LinkedList
{
    static Node head; // head of list
    boolean borrow;

    /* Node Class */
    static class Node {
        int data;
        Node next;
        
        // Constructor to create a new node
        Node(int d) {
            data = d;
            next = null;
        }
    }

    
    /* A utility function to get length of 
    linked list */
    int getLength(Node node) 
    {
        int size = 0;
        while (node != null)
        {
            node = node.next;
            size++;
        }
        return size;
    }

    /* A Utility that padds zeros in front 
    of the Node, with the given diff */
    Node paddZeros(Node sNode, int diff)
    {
        if (sNode == null)
            return null;

        Node zHead = new Node(0);
        diff--;
        Node temp = zHead;
        while ((diff--) != 0)
        {
            temp.next = new Node(0);
            temp = temp.next;
        }
        temp.next = sNode;
        return zHead;
    }

    /* Subtract LinkedList Helper is a recursive
    function, move till the last Node, and 
    subtract the digits and create the Node and
    return the Node. If d1 < d2, we borrow the 
    number from previous digit. */
    Node subtractLinkedListHelper(Node l1, Node l2)
    {
        if (l1 == null && l2 == null && borrow == false)
            return null;

        Node previous = subtractLinkedListHelper((l1 != null) ? 
                                 l1.next : null, (l2 != null) ? 
                                 l2.next : null);

        int d1 = l1.data;
        int d2 = l2.data;
        int sub = 0;

        /* if you have given the value value to 
        next digit then reduce the d1 by 1 */
        if (borrow)
        {
            d1--;
            borrow = false;
        }

        /* If d1 < d2 , then borrow the number from
        previous digit. Add 10 to d1 and set 
        borrow = true; */
        if (d1 < d2)
        {
            borrow = true;
            d1 = d1 + 10;
        }

        /* subtract the digits */
        sub = d1 - d2;

        /* Create a Node with sub value */
        Node current = new Node(sub);

        /* Set the Next pointer as Previous */
        current.next = previous;

        return current;
    }

    /* This API subtracts two linked lists and 
    returns the linked list which shall have the
    subtracted result. */
    Node subtractLinkedList(Node l1, Node l2)
    {
        // Base Case.
        if (l1 == null && l2 == null)
            return null;

        // In either of the case, get the lengths
        // of both Linked list.
        int len1 = getLength(l1);
        int len2 = getLength(l2);

        Node lNode = null, sNode = null;

        Node temp1 = l1;
        Node temp2 = l2;

        // If lengths differ, calculate the smaller 
        // Node and padd zeros for smaller Node and
        // ensure both larger Node and smaller Node
        // has equal length.
        if (len1 != len2)
        {
            lNode = len1 > len2 ? l1 : l2;
            sNode = len1 > len2 ? l2 : l1;
            sNode = paddZeros(sNode, Math.abs(len1 - len2));
        }

        else
        {
            // If both list lengths are equal, then 
            // calculate the larger and smaller list. 
            // If 5-6-7 & 5-6-8 are linked list, then 
            // walk through linked list at last Node 
            // as 7 < 8, larger Node is 5-6-8 and 
            // smaller Node is 5-6-7.
            while (l1 != null && l2 != null)
            {
                if (l1.data != l2.data)
                {
                    lNode = l1.data > l2.data ? temp1 : temp2;
                    sNode = l1.data > l2.data ? temp2 : temp1;
                    break;
                }
                l1 = l1.next;
                l2 = l2.next;
            }
        }

        // After calculating larger and smaller Node,
        // call subtractLinkedListHelper which returns
        // the subtracted linked list.
        borrow = false;
        return subtractLinkedListHelper(lNode, sNode);
    }

    // function to display the linked list
    static void printList(Node head)
    {
        Node temp = head;
        while (temp != null) 
        {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
    }

    // Driver program to test above
    public static void main (String[] args) 
    {
        Node head = new Node(1);
        head.next = new Node(0);
        head.next.next = new Node(0);

        Node head2 = new Node(1);
        
        LinkedList ob = new LinkedList();
        Node result = ob.subtractLinkedList(head, head2);

        printList(result);

    } 
}

// This article is contributed by Chhavi


Output :
0 9 9

This article is contributed by Mu Ven. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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