Compare numbers represented by Linked Lists

Given the pointers to the head nodes of two linked lists. The task is to compare the numbers represented by the linked lists. The numbers represented by the lists may contain leading zeros. 

1. If the numbers are equal then print 0.
2. If the number represented by the first linked list is greater then print 1.
3. If the number represented by the second linked list is greater then print -1.

Examples:

Input:
List1 = 2 -> 3 -> 1 -> 2 -> 4 -> NULL 
List2 = 2 -> 3 -> 2 -> 4 -> 2 -> NULL 
Output: -1

Input:
List1 = 0 -> 0 -> 1 -> 2 -> 4 -> NULL 
List2 = 0 -> 0 -> 0 -> 4 -> 2 -> NULL 
Output: 1 

Approach: Since the numbers may contain leading zeros, first remove all the leading zeros from the start of the linked lists. After that compare their lengths, if the lengths are unequal, this means that one of the numbers is definitely greater and returns 1 or -1 based upon whose length is greater. Else traverse both the lists simultaneously, and while traversing we compare the digits at every node. If at any point, the digit is unequal then return either 1 or -1 based on the value of the digits. If the end of the linked lists is reached then the linked lists are identical and hence return 0.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

/* Structure for a linked list node */
struct Node {
    int data;
    struct Node* next;
};

/* A helper function to remove zeros from
the start of the linked list */
Node* removeLeadingZeros(struct Node* a)
{
    if (a != NULL && a->data == 0)
        return removeLeadingZeros(a->next);
    else
        return a;
}

/* A helper function to find the length of
linked list*/
int getSize(struct Node* a)
{
    int sz = 0;
    while (a != NULL) {
        a = a->next;
        sz++;
    }
    return sz;
}

/* Given a reference (pointer to pointer) to the
head of a list and an int, push a new node on the
front of the list. */
void push(struct Node** head_ref, int new_data)
{
    /* Allocate node */
    struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));

    /* Set the data */
    new_node->data = new_data;

    /* Link the old list after the new node */
    new_node->next = (*head_ref);

    /* Set the head to point to the new node */
    (*head_ref) = new_node;
}

// Function to compare the numbers
// represented as linked lists
int compare(struct Node* a,
            struct Node* b)
{

    // Remover leading zeroes from
    // the linked lists
    a = removeLeadingZeros(a);
    b = removeLeadingZeros(b);

    int lenA = getSize(a);
    int lenB = getSize(b);

    if (lenA > lenB) {

        /* Since the number represented by a
        has a greater length, it will be greater*/
        return 1;
    }
    else if (lenB > lenA) {
        return -1;
    }

    /* If the lengths of two numbers are equal
        we have to check their magnitudes*/
    while (a != NULL && b != NULL) {
        if (a->data > b->data)
            return 1;
        else if (a->data < b->data)
            return -1;

        /* If we reach here, then a and b are
        not NULL and their data is same, so
        move to next nodes in both lists */
        a = a->next;
        b = b->next;
    }

    // If linked lists are identical, then
    // we need to return zero
    return 0;
}

// Driver code
int main()
{
    /* The constructed linked lists are :
    a: 5->6->7
    b: 2->3->3 */
    struct Node* a = NULL;
    push(&a, 7);
    push(&a, 6);
    push(&a, 5);

    struct Node* b = NULL;
    push(&b, 3);
    push(&b, 3);
    push(&b, 2);

    cout << compare(a, b);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

/* Structure for a linked list node */
static class Node
{
    int data;
    Node next;
};

/* A helper function to remove zeros from
the start of the linked list */
static Node removeLeadingZeros(Node a)
{
    if (a != null && a.data == 0)
        return removeLeadingZeros(a.next);
    else
        return a;
}

/* A helper function to find the length of
linked list*/
static int getSize(Node a)
{
    int sz = 0;
    while (a != null)
    {
        a = a.next;
        sz++;
    }
    return sz;
}

/* Given a reference (pointer to pointer) to the
head of a list and an int, push a new node on the
front of the list. */
static Node push(Node head_ref, int new_data)
{
    /* Allocate node */
    Node new_node = new Node();

    /* Set the data */
    new_node.data = new_data;

    /* Link the old list after the new node */
    new_node.next = head_ref;

    /* Set the head to point to the new node */
    head_ref = new_node;
    return head_ref;
}

// Function to compare the numbers
// represented as linked lists
static int compare(Node a,
                   Node b)
{

    // Remover leading zeroes from
    // the linked lists
    a = removeLeadingZeros(a);
    b = removeLeadingZeros(b);

    int lenA = getSize(a);
    int lenB = getSize(b);

    if (lenA > lenB)
    {

        /* Since the number represented by a
        has a greater length, it will be greater*/
        return 1;
    }

    else if (lenB > lenA)
    {
        return -1;
    }

    /* If the lengths of two numbers are equal
        we have to check their magnitudes*/
    while (a != null && b != null)
    {
        if (a.data > b.data)
            return 1;
        else if (a.data < b.data)
            return -1;

        /* If we reach here, then a and b are
        not null and their data is same, so
        move to next nodes in both lists */
        a = a.next;
        b = b.next;
    }

    // If linked lists are identical, then
    // we need to return zero
    return 0;
}

// Driver code
public static void main(String[] args)
{

    /* The constructed linked lists are :
    a: 5->6->7
    b: 2->3->3 */
    Node a = null;
    a = push(a, 7);
    a = push(a, 6);
    a = push(a, 5);

    Node b = null;
    b = push(b, 3);
    b = push(b, 3);
    b = push(b, 2);

    System.out.println(compare(a, b));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach

''' Structure for a linked list node '''
class Node:

    def __init__(self):
        self.data = 0
        self.next = None

''' A helper function to remove zeros from
the start of the linked list '''
def removeLeadingZeros(a):

    if (a != None and a.data == 0):
        return removeLeadingZeros(a.next);
    else:
        return a;

''' A helper function to find the length of
linked list'''
def getSize(a):

    sz = 0;

    while (a != None):
        a = a.next;
        sz += 1

    return sz;

''' Given a reference (pointer to pointer) to the
head of a list and an int, push a new node on the
front of the list. '''
def push(head_ref, new_data):

    ''' Allocate node '''
    new_node = Node()

    ''' Set the data '''
    new_node.data = new_data;

    ''' Link the old list after the new node '''
    new_node.next = (head_ref);

    ''' Set the head to point to the new node '''
    (head_ref) = new_node;

    return head_ref

# Function to compare the numbers
# represented as linked lists
def compare(a, b):

    # Remover leading zeroes from
    # the linked lists
    a = removeLeadingZeros(a);
    b = removeLeadingZeros(b);

    lenA = getSize(a);
    lenB = getSize(b);

    if (lenA > lenB):

        ''' Since the number represented by a
        has a greater length, it will be greater'''
        return 1;

    elif (lenB > lenA):
        return -1;

    ''' If the lengths of two numbers are equal
        we have to check their magnitudes'''
    while (a != None and b != None):
        if (a.data > b.data):
            return 1;
        elif (a.data < b.data):
            return -1;

        ''' If we reach here, then a and b are
        not None and their data is same, so
        move to next nodes in both lists '''
        a = a.next;
        b = b.next;

    # If linked lists are identical, then
    # we need to return zero
    return 0;

# Driver code
if __name__=='__main__':

    ''' The constructed linked lists are :
    a: 5.6.7
    b: 2.3.3 '''
    a = None;
    a = push(a, 7);
    a = push(a, 6);
    a = push(a, 5);

    b = None;
    b = push(b, 3);
    b = push(b, 3);
    b = push(b, 2);

    print(compare(a, b))

# This code is contributed by rutvik_56

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

/* Structure for a linked list node */
public class Node
{
    public int data;
    public Node next;
};

/* A helper function to remove zeros from
the start of the linked list */
static Node removeLeadingZeros(Node a)
{
    if (a != null && a.data == 0)
        return removeLeadingZeros(a.next);
    else
        return a;
}

/* A helper function to find the length of
linked list*/
static int getSize(Node a)
{
    int sz = 0;
    while (a != null)
    {
        a = a.next;
        sz++;
    }
    return sz;
}

/* Given a reference (pointer to pointer) to the
head of a list and an int, push a new node on the
front of the list. */
static Node push(Node head_ref, int new_data)
{
    /* Allocate node */
    Node new_node = new Node();

    /* Set the data */
    new_node.data = new_data;

    /* Link the old list after the new node */
    new_node.next = head_ref;

    /* Set the head to point to the new node */
    head_ref = new_node;
    return head_ref;
}

// Function to compare the numbers
// represented as linked lists
static int compare(Node a, Node b)
{

    // Remover leading zeroes from
    // the linked lists
    a = removeLeadingZeros(a);
    b = removeLeadingZeros(b);

    int lenA = getSize(a);
    int lenB = getSize(b);

    if (lenA > lenB)
    {

        /* Since the number represented by a
        has a greater length, it will be greater*/
        return 1;
    }

    else if (lenB > lenA)
    {
        return -1;
    }

    /* If the lengths of two numbers are equal
        we have to check their magnitudes*/
    while (a != null && b != null)
    {
        if (a.data > b.data)
            return 1;
        else if (a.data < b.data)
            return -1;

        /* If we reach here, then a and b are
        not null and their data is same, so
        move to next nodes in both lists */
        a = a.next;
        b = b.next;
    }

    // If linked lists are identical, then
    // we need to return zero
    return 0;
}

// Driver code
public static void Main(String[] args)
{

    /* The constructed linked lists are :
    a: 5->6->7
    b: 2->3->3 */
    Node a = null;
    a = push(a, 7);
    a = push(a, 6);
    a = push(a, 5);

    Node b = null;
    b = push(b, 3);
    b = push(b, 3);
    b = push(b, 2);

    Console.WriteLine(compare(a, b));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript implementation of the approach

/* Structure for a linked list node */
class Node {
        constructor() {
                this.data = 0;
                this.next = null;
             }
        }

/* A helper function to remove zeros from
the start of the linked list */
function removeLeadingZeros( a)
{
    if (a != null && a.data == 0)
        return removeLeadingZeros(a.next);
    else
        return a;
}

/* A helper function to find the length of
linked list*/
function getSize( a)
{
    let sz = 0;
    while (a != null)
    {
        a = a.next;
        sz++;
    }
    return sz;
}

/* Given a reference (pointer to pointer) to the
head of a list and an int, push a new node on the
front of the list. */
function push( head_ref,  new_data)
{
    /* Allocate node */
    var new_node = new Node();

    /* Set the data */
    new_node.data = new_data;

    /* Link the old list after the new node */
    new_node.next = head_ref;

    /* Set the head to point to the new node */
    head_ref = new_node;
    return head_ref;
}

// Function to compare the numbers
// represented as linked lists
function compare( a, b)
{

    // Remover leading zeroes from
    // the linked lists
    a = removeLeadingZeros(a);
    b = removeLeadingZeros(b);

    let lenA = getSize(a);
    let lenB = getSize(b);

    if (lenA > lenB)
    {

        /* Since the number represented by a
        has a greater length, it will be greater*/
        return 1;
    }

    else if (lenB > lenA)
    {
        return -1;
    }

    /* If the lengths of two numbers are equal
        we have to check their magnitudes*/
    while (a != null && b != null)
    {
        if (a.data > b.data)
            return 1;
        else if (a.data < b.data)
            return -1;

        /* If we reach here, then a and b are
        not null and their data is same, so
        move to next nodes in both lists */
        a = a.next;
        b = b.next;
    }

    // If linked lists are identical, then
    // we need to return zero
    return 0;
}


// Driver Code

/* The constructed linked lists are :
a: 5->6->7
b: 2->3->3 */
var a = null;
a = push(a, 7);
a = push(a, 6);
a = push(a, 5);

var b = null;
b = push(b, 3);
b = push(b, 3);
b = push(b, 2);

document.write(compare(a, b));

</script>

Output:

1

Time Complexity: O(max(N, M)) where N and M are the lengths of the linked lists.

Auxiliary Space: O(1)