# First common element in two linked lists

• Difficulty Level : Basic
• Last Updated : 03 Aug, 2022

Given two Linked Lists, find the first common element between given linked list i.e., we need to find first node of the first list which is also present in the second list.

Examples:

```Input :
List1: 10->15->4->20
Lsit2:  8->4->2->10
Output : 10

Input :
List1: 1->2->3->4
Lsit2:  5->6->3->8
Output : 3```

We traverse the first list and for every node, we search it in the second list. As soon as we find an element in the second list, we return it.

## C++

 `// C++ program to find first common element in``// two unsorted linked list``#include ``using` `namespace` `std;` `/* Link list node */``struct` `Node {``    ``int` `data;``    ``struct` `Node* next;``};` `/* A utility function to insert a node at the``   ``beginning of a linked list*/``void` `push(``struct` `Node** head_ref, ``int` `new_data)``{``    ``struct` `Node* new_node =``          ``(``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));``    ``new_node->data = new_data;``    ``new_node->next = (*head_ref);``    ``(*head_ref) = new_node;``}` `/* Returns the first repeating element in linked list*/``int` `firstCommon(``struct` `Node* head1, ``struct` `Node* head2)``{``    ``// Traverse through every node of first list``    ``for` `(; head1 != NULL; head1=head1->next)` `       ``// If current node is present in second list``       ``for` `(Node *p = head2; p != NULL; p = p->next)``            ``if` `(p->data == head1->data)``                ``return` `head1->data;` `    ``// If no common node``    ``return` `0;``}` `// Driver code``int` `main()``{``    ``struct` `Node* head1 = NULL;``    ``push(&head1, 20);``    ``push(&head1, 5);``    ``push(&head1, 15);``    ``push(&head1, 10);` `    ``struct` `Node* head2 = NULL;``    ``push(&head2, 10);``    ``push(&head2, 2);``    ``push(&head2, 15);``    ``push(&head2, 8);` `    ``cout << firstCommon(head1, head2);``    ``return` `0;``}`

## Java

 `// Java program to find first common element``// in two unsorted linked list``import` `java.util.*;` `class` `GFG``{` `/* Link list node */``static` `class` `Node``{``    ``int` `data;``    ``Node next;``};` `/* A utility function to insert a node at the``beginning of a linked list*/``static` `Node push(Node head_ref, ``int` `new_data)``{``    ``Node new_node = ``new` `Node();``    ``new_node.data = new_data;``    ``new_node.next = head_ref;``    ``head_ref = new_node;``    ``return` `head_ref;``}` `/* Returns the first repeating element``   ``in linked list*/``static` `int` `firstCommon(Node head1, Node head2)``{``    ``// Traverse through every node of first list``    ``for` `(; head1 != ``null``; head1 = head1.next)` `    ``// If current node is present in second list``    ``for` `(Node p = head2; p != ``null``; p = p.next)``            ``if` `(p.data == head1.data)``                ``return` `head1.data;` `    ``// If no common node``    ``return` `0``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``Node head1 = ``null``;``    ``head1 = push(head1, ``20``);``    ``head1 = push(head1, ``5``);``    ``head1 = push(head1, ``15``);``    ``head1 = push(head1, ``10``);` `    ``Node head2 = ``null``;``    ``head2 = push(head2, ``10``);``    ``head2 = push(head2, ``2``);``    ``head2 = push(head2, ``15``);``    ``head2 = push(head2, ``8``);` `    ``System.out.println(firstCommon(head1, head2));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to find first common element``# in two unsorted linked list``import` `math` `# Link list node``class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.``next` `=` `None` `# A utility function to insert a node at the``#beginning of a linked list``def` `push(head_ref, new_data):``    ``new_node ``=` `Node(new_data)``    ``new_node.data ``=` `new_data``    ``new_node.``next` `=` `head_ref``    ``head_ref ``=` `new_node``    ``return` `head_ref` `# Returns the first repeating element``# in linked list*/``def` `firstCommon(head1, head2):``    ` `    ``# Traverse through every node of first list``    ``while``(head1 !``=` `None``):``        ``p ``=` `head2` `    ``# If current node is present in second list``        ``while``(p !``=` `None``):``            ``if` `(p.data ``=``=` `head1.data):``                ``return` `head1.data``            ``p ``=` `p.``next``        ``head1 ``=` `head1.``next``        ` `    ``# If no common node``    ``return` `0` `# Driver code``if` `__name__``=``=``'__main__'``:``    ``head1 ``=` `None``    ``head1 ``=` `push(head1, ``20``)``    ``head1 ``=` `push(head1, ``5``)``    ``head1 ``=` `push(head1, ``15``)``    ``head1 ``=` `push(head1, ``10``)` `    ``head2 ``=` `None``    ``head2 ``=` `push(head2, ``10``)``    ``head2 ``=` `push(head2, ``2``)``    ``head2 ``=` `push(head2, ``15``)``    ``head2 ``=` `push(head2, ``8``)` `    ``print``(firstCommon(head1, head2))` `# This code is contributed by Srathore`

## C#

 `// C# program to find first common element``// in two unsorted linked list``using` `System;``using` `System.Collections.Generic;``    ` `class` `GFG``{` `/* Link list node */``class` `Node``{``    ``public` `int` `data;``    ``public` `Node next;``};` `/* A utility function to insert a node at the``beginning of a linked list*/``static` `Node push(Node head_ref, ``int` `new_data)``{``    ``Node new_node = ``new` `Node();``    ``new_node.data = new_data;``    ``new_node.next = head_ref;``    ``head_ref = new_node;``    ``return` `head_ref;``}` `/* Returns the first repeating element``in linked list*/``static` `int` `firstCommon(Node head1, Node head2)``{``    ``// Traverse through every node of first list``    ``for` `(; head1 != ``null``; head1 = head1.next)` `    ``// If current node is present in second list``    ``for` `(Node p = head2; p != ``null``; p = p.next)``            ``if` `(p.data == head1.data)``                ``return` `head1.data;` `    ``// If no common node``    ``return` `0;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``Node head1 = ``null``;``    ``head1 = push(head1, 20);``    ``head1 = push(head1, 5);``    ``head1 = push(head1, 15);``    ``head1 = push(head1, 10);` `    ``Node head2 = ``null``;``    ``head2 = push(head2, 10);``    ``head2 = push(head2, 2);``    ``head2 = push(head2, 15);``    ``head2 = push(head2, 8);` `    ``Console.WriteLine(firstCommon(head1, head2));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`10`

Time complexity: O(M*N) where M and N are size of given linked lists

Auxiliary Space: O(1)

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