Finding the intersection and union of 2 singly linked list - c

Question

The program generates the union and intersection of 2 linked lists. However, it does not generate the expected output where duplicate values from list 1 and list 2 should not be present and the intersection list is not even being displayed.
How can I resolve this issue?
`
#include<stdio.h>
#include<stdlib.h>
typedef struct node{
int data;
struct node *next;
} node;
void insertAtBeg(node **head, int ele){
node *newnode = (node*)malloc(sizeof(node));
newnode->data = ele;
newnode->next = (*head);
(*head) = newnode;
}
int isPresent(node *temp, int ele){
node *t = temp;
while(t != NULL){
if(t->data == ele)
return 1;
t = t->next;
}
return 0;
}
void printList(node *n){
while(n != NULL){
printf("%d->",n->data);
n = n->next;
}
}
node* getUnion(node *head1,node *head2){
node *result = NULL;
node *t1 = head1;
node *t2 = head2;
while(t1 != NULL){
insertAtBeg(&result, t1->data);
t1 = t1->next;
}
while(t2!=NULL){
if(!isPresent(result, t2->data));
insertAtBeg(&result,t2->data);
t2 = t2->next;
}
return result;
}
node *getIntersection(node *head1,node *head2){
node* result = NULL;
node* t1 = head1;
while(t1 != NULL){
if(isPresent(head2, t1->data))
insertAtBeg(&result,t1->data);
t1 = t1->next;
}
return result;
}
int main(){
node *intersection = NULL;
node *unin = NULL;
node *List1;
node *List2;
List1 = List2 = NULL;
int i,n,m,temp;
printf("Enter the size of the first linked list:\n");
scanf("%d",&n);
printf("Enter %d elements\n",n);
for(i = 0;i < n;i++){
scanf("%d",&temp);
insertAtBeg(&List1,temp);
}
printf("Displaying list 1:\n");
printList(List1);
printf("\nEnter the size of the second linked list:\n");
scanf("%d",&m);
printf("Enter %d elements\n",m);
for(i = 0;i < m;i++){
scanf("%d",&temp);
insertAtBeg(&List2,temp);
}
printf("Displaying list 2:\n");
printList(List2);
unin = getUnion(List1,List2);
intersection = getIntersection(List1,List2);
printf("\nLinked List with Union of List1 and List2:\n");
printList(unin);
printf("\nLinked List with Intersection of List1 and List2:\n");
printList(intersection);
return 0;
}
`
Generated output:
Enter the size of the first linked list:
4
Enter 4 elements
3 4 6 1
Displaying list 1:
1->6->4->3->
Enter the size of the second linked list:
4
Enter 4 elements
8 9 10 1
Displaying list 2:
1->10->9->8->
Linked List with Union of List1 and List2:
8->9->10->1->3->4->6->1->
Linked List with Intersection of List1 and List2:
1->

Answer 1

Within the function getUnion there is a typo
while(t2!=NULL){
if(!isPresent(result, t2->data));
^^^^
insertAtBeg(&result,t2->data);
t2 = t2->next;
}
Remove the semicolon.
Also after this call
printList(intersection);
output the new line character like for example
putchar( '\n' );
Pay attention to that you should declare parameters that denotes pointers to nodes with qualifier const if lists are not changed within functions as for example
int isPresent( const node *temp, int ele){
const node *t = temp;
while(t != NULL){
if(t->data == ele)
return 1;
t = t->next;
}
return 0;
}
void printList(const node *n){
while(n != NULL){
printf("%d->",n->data);
n = n->next;
}
}
node* getUnion( const node *head1,const node *head2){
node *result = NULL;
const node *t1 = head1;
const node *t2 = head2;
while(t1 != NULL){
insertAtBeg(&result, t1->data);
t1 = t1->next;
}
while(t2!=NULL){
if(!isPresent(result, t2->data))
insertAtBeg(&result,t2->data);
t2 = t2->next;
}
return result;
}
node *getIntersection(const node *head1,const node *head2){
node* result = NULL;
const node* t1 = head1;
while(t1 != NULL){
if(isPresent(head2, t1->data))
insertAtBeg(&result,t1->data);
t1 = t1->next;
}
return result;
}