Question

Count all nonzero elements, odd numbers and even numbers in a linked list.

Code needed in java.

Answer 1

#include<stdio.h>
#include<conio.h>
struct node
{
int data;
struct node *link;
}*start;
void create(int);
void disp();
void count();
void main()
{
int ch,n,i,m,a,pos;
clrscr();
start=NULL;
do
{
printf(“\n\nMENU\n\n”);
printf(“\n1.CREATE\n”);
printf(“\n2.DISPLAY\n”);
printf(“\n3.COUNT\n”);
printf(“\n4.EXIT\n”);
printf(“\nENTER UR CHOICE\n”);
scanf(“%d”,&ch);
switch(ch)
{
case 1:
printf(“\n\nHOW MANY NODES U WANT TO CREATE\n”);
scanf(“%d”,&n);
for(i=0;i<n;i++)
{
printf(“\nENTER THE DATA”);
scanf(“%d”,&m);
create(m);
}
break;
case 3:
count();
break;
case 2:
disp();
break;
case 4:
exit(0);
}
}
while(ch!=4);
getch();
}
void count()
{
struct node *q;
int nonz=0,eno=0,ono=0;
q=start;
while(q!=NULL)
{
if(q->data>0)
{
nonz++;
}
if(q->data%2==0)
{
eno++;
}
else
{
ono++;
}

q=q->link;
}
printf(“\n\nPOSITVE NO ARE %d EVEN NO ARE %d ODD NO ARE %d”,nonz,eno,ono);
}

void create(int data)
{
struct node *q,*tmp;
tmp=(struct node *)malloc(sizeof(struct node));
tmp->data=data;
tmp->link=NULL;
if(start==NULL)
{
start=tmp;
}
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}
void disp()
{
struct node *q;
if(start==NULL)
{
printf(“\n\nLIST IS EMPTY”);
}
else
{
q=start;
while(q!=NULL)
{
printf(“%d->”,q->data);
q=q->link;
}
printf(“NULL”);
}
}