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Java Program To Subtract Two Numbers Represented As Linked Lists

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  • Last Updated : 03 Jan, 2022

Given two linked lists that represent two large positive numbers. Subtract the smaller number from the 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 the larger ones.
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
Explanation: Number represented as 
lists are 100 and 1, so 100 - 1 is 099

Input: l1 = 7-> 8 -> 6 -> NULL,  l2 = 7 -> 8 -> 9 NULL
Output: 3->NULL
Explanation: Number represented as 
lists are 786 and  789, so 789 - 786 is 3, 
as the smaller value is subtracted from 
the larger one.

Approach: Following are the steps.

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

Following is the implementation of the above approach.

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 
{
    // Head of list
    static Node head; 
    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 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 code
    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 

Complexity Analysis:

  • Time complexity: O(n). 
    As no nested traversal of linked list is needed.
  • Auxiliary Space: O(n). 
    If recursive stack space is taken into consideration O(n) space is needed.

Please refer complete article on Subtract Two Numbers represented as Linked Lists for more details!


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