Identical Linked Lists

Difficulty Level : Basic
Last Updated : 12 May, 2021

Two Linked Lists are identical when they have the same data and the arrangement of data is also the same. For example, Linked lists a (1->2->3) and b(1->2->3) are identical. . Write a function to check if the given two linked lists are identical.

Method 1 (Iterative)
To identify if two lists are identical, we need to traverse both lists simultaneously, and while traversing we need to compare data.

C++

// An iterative C++ program to check if
// two linked lists are identical or not
#include<bits/stdc++.h>
using namespace std;
 
/* Structure for a linked list node */
struct Node
{
    int data;
    struct Node *next;
};
 
/* Returns true if linked lists a and b
are identical, otherwise false */
bool areIdentical(struct Node *a,
                  struct Node *b)
{
    while (a != NULL && b != NULL)
    {
        if (a->data != b->data)
            return false;
 
        /* 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
    // 'a' and 'b' must be NULL at this point.
    return (a == NULL && b == NULL);
}
 
/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
/* 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));
 
    /* 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;
}
 
// Driver Code
int main()
{
    /* The constructed linked lists are :
    a: 3->2->1
    b: 3->2->1 */
    struct Node *a = NULL;
    struct Node *b = NULL;
    push(&a;, 1);
    push(&a;, 2);
    push(&a;, 3);
    push(&b;, 1);
    push(&b;, 2);
    push(&b;, 3);
 
    if(areIdentical(a, b))
        cout << "Identical";
    else
        cout << "Not identical";
 
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

// An iterative C program to check if two linked lists are
// identical or not
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
 
/* Structure for a linked list node */
struct Node
{
    int data;
    struct Node *next;
};
 
/* Returns true if linked lists a and b are identical,
   otherwise false */
bool areIdentical(struct Node *a, struct Node *b)
{
    while (a != NULL && b != NULL)
    {
        if (a->data != b->data)
            return false;
 
        /* 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 'a' and 'b' must
    // be NULL at this point.
    return (a == NULL && b == NULL);
}
 
/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
/* 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));
 
    /* 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;
}
 
/* Driver program to test above function */
int main()
{
    /* The constructed linked lists are :
     a: 3->2->1
     b: 3->2->1 */
    struct Node *a = NULL;
    struct Node *b = NULL;
    push(&a;, 1);
    push(&a;, 2);
    push(&a;, 3);
    push(&b;, 1);
    push(&b;, 2);
    push(&b;, 3);
 
    areIdentical(a, b)? printf("Identical"):
                        printf("Not identical");
 
    return 0;
}

Java

// An iterative Java program to check if two linked lists
// are identical or not
class LinkedList
{
    Node head;  // head of list
 
    /* Linked list Node*/
    class Node
    {
        int data;
        Node next;
        Node(int d) { data = d; next = null; }
    }
 
    /* Returns true if linked lists a and b are identical,
       otherwise false */
    boolean areIdentical(LinkedList listb)
    {
        Node a = this.head, b = listb.head;
        while (a != null && b != null)
        {
            if (a.data != b.data)
                return false;
 
            /* 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 'a' and 'b' must
        // be null at this point.
        return (a == null && b == null);
    }
 
    /* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
    /*  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(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);
 
        /* 3. Make next of new Node as head */
        new_node.next = head;
 
        /* 4. Move the head to point to new Node */
        head = new_node;
    }
 
 
    /* Driver program to test above functions */
    public static void main(String args[])
    {
        LinkedList llist1 = new LinkedList();
        LinkedList llist2 = new LinkedList();
 
        /* The constructed linked lists are :
           llist1: 3->2->1
           llist2: 3->2->1 */
 
        llist1.push(1);
        llist1.push(2);
        llist1.push(3);
 
        llist2.push(1);
        llist2.push(2);
        llist2.push(3);
 
        if (llist1.areIdentical(llist2) == true)
            System.out.println("Identical ");
        else
            System.out.println("Not identical ");
 
    }
} /* This code is contributed by Rajat Mishra */

Python3

# An iterative Java program to check if
# two linked lists are identical or not
 
# Linked list Node
class Node:
    def __init__(self, d):
        self.data = d
        self.next = None
 
class LinkedList:
    def __init__(self):
        self.head = None # head of list
     
    # Returns true if linked lists a and b
    # are identical, otherwise false
    def areIdentical(self, listb):
        a = self.head
        b = listb.head
        while (a != None and b != None):
            if (a.data != b.data):
                return False
 
            # 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 'a' and 'b' must be null
        # at this point.
        return (a == None and b == None)
 
    # UTILITY FUNCTIONS TO TEST fun1() and fun2()
    # 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(self, new_data):
         
        # 1 & 2: Allocate the Node &
        # Put in the data
        new_node = Node(new_data)
 
        # 3. Make next of new Node as head
        new_node.next = self.head
 
        # 4. Move the head to point to new Node
        self.head = new_node
 
# Driver Code
llist1 = LinkedList()
llist2 = LinkedList()
 
# The constructed linked lists are :
# llist1: 3->2->1
# llist2: 3->2->1
llist1.push(1)
llist1.push(2)
llist1.push(3)
llist2.push(1)
llist2.push(2)
llist2.push(3)
 
if (llist1.areIdentical(llist2) == True):
    print("Identical ")
else:
    print("Not identical ")
 
# This code is contributed by Prerna Saini

C#

// An iterative C# program to
// check if two linked lists
// are identical or not
using System;
     
public class LinkedList
{
    Node head; // head of list
 
    /* Linked list Node*/
    public class Node
    {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d; next = null;
        }
    }
 
    /* Returns true if linked lists
    a and b are identical,
    otherwise false */
    bool areIdentical(LinkedList listb)
    {
        Node a = this.head, b = listb.head;
        while (a != null && b != null)
        {
            if (a.data != b.data)
                return false;
 
            /* 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 'a' and 'b' must
        // be null at this point.
        return (a == null && b == null);
    }
 
    /* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
    /* 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(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                Put in the data*/
        Node new_node = new Node(new_data);
 
        /* 3. Make next of new Node as head */
        new_node.next = head;
 
        /* 4. Move the head to point to new Node */
        head = new_node;
    }
 
 
    /* Driver code */
    public static void Main(String []args)
    {
        LinkedList llist1 = new LinkedList();
        LinkedList llist2 = new LinkedList();
 
        /* The constructed linked lists are :
        llist1: 3->2->1
        llist2: 3->2->1 */
 
        llist1.push(1);
        llist1.push(2);
        llist1.push(3);
 
        llist2.push(1);
        llist2.push(2);
        llist2.push(3);
 
        if (llist1.areIdentical(llist2) == true)
            Console.WriteLine("Identical ");
        else
            Console.WriteLine("Not identical ");
    }
}
 
// This code contributed by Rajput-Ji

Output:

Identical

Method 2 (Recursive)
Recursive solution code is much cleaner than iterative code. You probably wouldn’t want to use the recursive version for production code, however, because it will use stack space which is proportional to the length of the lists

C++

// A recursive C++ function to check if two linked
// lists are identical or not
bool areIdentical(Node *a, Node *b)
{
    // If both lists are empty
    if (a == NULL && b == NULL)
    return true;
 
    // If both lists are not empty, then data of
    // current nodes must match, and same should
    // be recursively true for rest of the nodes.
    if (a != NULL && b != NULL)
    return (a->data == b->data) &&
            areIdentical(a->next, b->next);
 
    // If we reach here, then one of the lists
    // is empty and other is not
    return false;
}
 
//This is code is contributed by rathbhupendra

C

// A recursive C function to check if two linked
// lists are identical or not
bool areIdentical(struct Node *a, struct Node *b)
{
    // If both lists are empty
    if (a == NULL && b == NULL)
       return true;
 
    // If both lists are not empty, then data of
    // current nodes must match, and same should
    // be recursively true for rest of the nodes.
    if (a != NULL && b != NULL)
       return (a->data == b->data) &&
              areIdentical(a->next, b->next);
 
    // If we reach here, then one of the lists
    // is empty and other is not
    return false;
}

Java

// A recursive Java method to check if two linked
// lists are identical or not
boolean areIdenticalRecur(Node a, Node b)
{
    // If both lists are empty
    if (a == null && b == null)
        return true;
 
    // If both lists are not empty, then data of
    // current nodes must match, and same should
    // be recursively true for rest of the nodes.
    if (a != null && b != null)
        return (a.data == b.data) &&
               areIdenticalRecur(a.next, b.next);
 
    // If we reach here, then one of the lists
    // is empty and other is not
    return false;
}
 
/* Returns true if linked lists a and b are identical,
   otherwise false */
boolean areIdentical(LinkedList listb)
{
    return areIdenticalRecur(this.head, listb.head);
}

Python3

# A recursive Python3 function to check
# if two linked lists are identical or not
def areIdentical(a, b):
     
    # If both lists are empty
    if (a == None and b == None):
        return True
 
    # If both lists are not empty,
    # then data of current nodes must match,
    # and same should be recursively true
    # for rest of the nodes.
    if (a != None and b != None):
        return ((a.data == b.data) and
                 areIdentical(a.next, b.next))
 
    # If we reach here, then one of the lists
    # is empty and other is not
    return False
 
# This code is contributed by Princi Singh

C#

// A recursive C# method to
// check if two linked lists
// are identical or not
bool areIdenticalRecur(Node a, Node b)
{
    // If both lists are empty
    if (a == null && b == null)
        return true;
 
    // If both lists are not empty, then data of
    // current nodes must match, and same should
    // be recursively true for rest of the nodes.
    if (a != null && b != null)
        return (a.data == b.data) &&
            areIdenticalRecur(a.next, b.next);
 
    // If we reach here, then one of the lists
    // is empty and other is not
    return false;
}
 
/* Returns true if linked lists
a and b are identical, otherwise false */
bool areIdentical(LinkedList listb)
{
    return areIdenticalRecur(this.head, listb.head);
}
}
 
// This code is contributed by princiraj1992

Javascript

<script>
// A recursive javascript method to check if two linked
// lists are identical or not
function areIdenticalRecur( a,  b)
{
    // If both lists are empty
    if (a == null && b == null)
        return true;
 
    // If both lists are not empty, then data of
    // current nodes must match, and same should
    // be recursively true for rest of the nodes.
    if (a != null && b != null)
        return (a.data == b.data) &&
               areIdenticalRecur(a.next, b.next);
 
    // If we reach here, then one of the lists
    // is empty and other is not
    return false;
}
 
/* Returns true if linked lists a and b are identical,
   otherwise false */
function areIdentical( listb)
{
    return areIdenticalRecur(this.head, listb.head);
}
 
// This code contributed by aashish1995
</script>

Time Complexity: O(n) for both iterative and recursive versions. n is the length of the smaller list among a and b.

?list=PLqM7alHXFySH41ZxzrPNj2pAYPOI8ITe7
Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.

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