Intersection of two sorted Linked lists

PROBLEM :

Given two lists sorted in increasing order, create a new list representing the intersection of the two lists. The new list should be made with its own memory — the original lists should not be changed.

For example, let the first linked list be 1->2->3->4->6 and second linked list be 2->4->6->8, then your function should create a third list as 2->4->6.


Input:

You have to complete the method which takes 3 argument: the head of the first linked list , the head of second linked list and the head of the third link list which is to be created. You should not read any input from stdin/console. There are multiple test cases. For each test case, this method will be called individually.

Output:
Complete the Function given to produce the desired list with intersectioned values.

Constraints:
1 <=T<= 20
1 <= size of linked lists <= 30
1 <= Data in linked list nodes <= 30


Example:(The input/output format below should be used for Expected Output only)

Input
1                       -->  (No of test cases)
5 4                     -->  (sizes of linked lists)
1 2 3 4 6               -->  (Elements of 1st linked list)
2 4 6 8                 -->  (Elements of 2nd linked list)

Output
2 4 6                   -->  (Elements of resultant 3rd linked list


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SIMPLE c++ IMPLEMENTATION :
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/* The structure of the Linked list Node is as follows:
struct node {
    int val;
    struct node* next;
}; */

void intersection(struct node **head1, struct node **head2, struct node **head3)
{
    struct node *one,*two,*temp,*ptr ;
    one=(*head1) ;
    two=(*head2) ;
   
    while(one!=NULL&&two!=NULL)
    {
        temp=(struct node*)malloc(sizeof(struct node)) ;
        temp->next=NULL ;
       
        if(one->val>two->val)
            two=two->next ;
        else if(one->val<two->val)
            one=one->next ;
        else
            {
                temp->val=one->val ;
                one=one->next ;
                two=two->next ;
               
                if((*head3)==NULL)
                    (*head3)=temp ;
                else
                {
                    ptr=(*head3) ;
                    while(ptr->next!=NULL)
                        ptr=ptr->next ;
                    ptr->next=temp ;
                }
            }
    }
}

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