Two Linked Lists are identical when they have same data and arrangement of data is also 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.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 ` `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 ` `#include ` `#include ` ` `  `/* 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 the 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); ` `} `

## 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 `

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

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

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :

4

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.