First common element in two linked lists

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 <bits/stdc++.h>
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

<script>

// Javascript program to find first common element
// in two unsorted linked list

/* Link list node */
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
};

/* A utility function to insert a node at the
beginning of a linked list*/
function push(head_ref, new_data)
{
    var 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*/
function firstCommon(head1, head2)
{
    // Traverse through every node of first list
    for (; head1 != null; head1 = head1.next)

    // If current node is present in second list
    for (var p = head2; p != null; p = p.next)
            if (p.data == head1.data)
                return head1.data;

    // If no common node
    return 0;
}

// Driver code
var head1 = null;
head1 = push(head1, 20);
head1 = push(head1, 5);
head1 = push(head1, 15);
head1 = push(head1, 10);
var head2 = null;
head2 = push(head2, 10);
head2 = push(head2, 2);
head2 = push(head2, 15);
head2 = push(head2, 8);
document.write(firstCommon(head1, head2));


</script>

Output:

10

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

Auxiliary Space: O(1)