union of linked lists - c

Question

I have created a program to find union of 2 linked list. My logic is first of all take a new list insert list1 contents into this list and insert only those values from list2 which are not in result list. My code is:
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
/* Linked list node */
struct node
{
int data;
struct node* next;
};
struct node* result;
struct node *newNode(int data)
{
struct node *new_node = (struct node *) malloc(sizeof(struct node));
new_node->data = data;
new_node->next = NULL;
return new_node;
}
/* Function to insert a node at the beginning of the Doubly Linked List */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node = newNode(new_data);
new_node->next=*head_ref;
*head_ref=new_node;
}
struct node *union3(struct node *first, struct node *second, struct node *result)
{
int flag = 0;
struct node *temp = NULL;
temp = first;
struct node *temp2 = NULL;
temp2 = second;
int value;
while (temp != NULL)
{
push(&result, temp->data); // pushing first list values in result
temp = temp->next;
}
while (second)
{
present(second->data, result); // comparing second list each member with result
second = second->next;
}
return result;
}
void present(int data, struct node *result1)
{
printf("The value in the beginning of present function is:");
int flag = 0;
while (result1)
{
if (result1->data == data)
{
flag++;
}
result1 = result1->next;
}
if (flag > 0)
{
printf("");
}
else
{
push(&result, data);
}
}
void printList(struct node *node)
{
while(node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
printf("\n");
}
/* Drier program to test above function */
int main(void)
{
struct node* first = NULL;
struct node* second=NULL;
// struct node* result=NULL;
struct node* union2=NULL;
// create first list 7->5->9->4->6
push(&first, 6);
push(&first, 4);
push(&first, 9);
push(&first, 5);
push(&first, 7);
printf("First list is:");
printList(first);
push(&second,6);
push(&second,4);
push(&second,9);
push(&second,11);
push(&second,12);
printf("second list is");
printList(second);
printf("their union is:");
union2=union3(first,second,result);
printf("Union of 2 lists is:");
printList(union2);
return 0;
}
Basically my logic is right but a problem is coming in result variable. Its list1 values pushed in it gets lost in it when it goes in present() function even though I have made result a global variable. Can anybody tell why output is displaying only list1 contents as:
output:6 4 9 5 7

Answer 1

With your algorithm if list1 has duplicates they will show in the final result, but if list2 has duplicates they will not show in the final result which is something you probably don't want.
Also I think you meant to use temp2 instead of second in:
while(second)
{
present(second->data,result); //comparing second list each member with result
second=second->next;
}
and finally this took me some time but I found your error:
push(&result,data);
should be result1
Hope this helps!

Answer 2

Catch! :) All you else need is to add a function that will free allocated memory for the lists.
#include <stdlib.h>
#include <stdio.h>
/* Linked list node */
struct node
{
int data;
struct node *next;
};
struct node *newNode( int data, struct node *next )
{
struct node *tmp = ( struct node *) malloc( sizeof( struct node ) );
if ( tmp )
{
tmp->data = data;
tmp->next = next;
}
return tmp;
}
/* Function to insert a node at the beginning of the Doubly Linked List */
void push( struct node **head_ref, int data )
{
/* allocate node */
struct node *tmp = newNode( data, *head_ref );
if ( tmp != NULL )
{
*head_ref = tmp;
}
}
int find( struct node *head, int data )
{
while ( head && head->data != data ) head = head->next;
return head != NULL;
}
struct node* list_union( struct node *first, struct node *second )
{
struct node *head = NULL;
struct node **head_ref = &head;
for ( struct node *current = first; current != NULL; current = current->next )
{
struct node *tmp = newNode( current->data, NULL );
if ( tmp != NULL )
{
*head_ref = tmp;
head_ref = &( *head_ref )->next;
}
}
for ( struct node *current = second; current != NULL; current = current->next )
{
if ( !find( first, current->data ) )
{
struct node *tmp = newNode( current->data, NULL );
if ( tmp != NULL )
{
*head_ref = tmp;
head_ref = &( *head_ref )->next;
}
}
}
return head;
}
void printList( struct node *node )
{
for ( ; node != NULL; node = node->next )
{
printf( "%d ", node->data );
}
printf("\n");
}
/* Drier program to test above function */
int main( void )
{
struct node *first = NULL;
struct node *second = NULL;
struct node *union2 = NULL;
// create first list 7->5->9->4->6
push( &first, 6 );
push( &first, 4 );
push( &first, 9 );
push( &first, 5 );
push( &first, 7 );
printf( "First list is: " );
printList( first );
push( &second, 6 );
push( &second, 4 );
push( &second, 9 );
push( &second, 11 );
push( &second, 12 );
printf( "second list is: " );
printList( second );
union2 = list_union( first, second );
printf( "Union of 2 lists is: " );
printList( union2 );
// Here you should call a function that frees all the memory allocated for the lists
}
The program output is
First list is: 7 5 9 4 6
second list is: 12 11 9 4 6
Union of 2 lists is: 7 5 9 4 6 12 11
Also you could write function push such a way that it would not add duplicated values to the list.