Queue

Posted by prawin
-
0 comments

Queue : queue means first in first out.(FIFO)

#include<stdio.h>
#define max 3
struct node
{
        int a[max];
        int first;
        int last;
};
void init(struct node*);
void push(struct node*,int);
int pop(struct node*);

main()
{
        struct node s;
        int i;
        init(&s);
        push(&s,20);
        push(&s,33);
        push(&s,12);
        push(&s,50);
        i = pop(&s);
        printf("the deleted number is %d\n",i);
}
void init(struct node *q)
{
        q->first = -1;
        q->last = -1;
}
void push(struct node *q,int num)
{
        if(q->last == max-1)
        {
        printf("que is full\n");
        }
        else
        {
        q->last++;
        q->a[q->last] = num;
        }
        if(q->first == -1)
        q->first++;
}
int pop(struct node *q)
{
        int n;
        if(q->last == -1)
        {
        printf("que is empty\n");
        }
        else
        {
        n = q->a[q->first];
        q->first++;
        return n;
        }
}
op :
que is full
the deleted number is 20

Author

Written by Admin

Aliquam molestie ligula vitae nunc lobortis dictum varius tellus porttitor. Suspendisse vehicula diam a ligula malesuada a pellentesque turpis facilisis. Vestibulum a urna elit. Nulla bibendum dolor suscipit tortor euismod eu laoreet odio facilisis.

0 comments:

Subscribe to: Post Comments (Atom)

About

#include
struct node
{
int data;
struct node* next;
};
void addatbeg(struct node,int);
void append(struct node,int);
void display(struct node*);
void delete(struct node**,int num);
void count(struct node);
void insert(struct node,int,int);
void reverse(struct node);
void reversing(struct node);
void mid(struct node);
main()
{
struct node head;
head = NULL;
addatbeg(&head,10);
addatbeg(&head,20);
addatbeg(&head,30);
display(head);
append(&head,60);
append(&head,70);
printf("\n");
display(head);
delete(&head,60);
printf("\n");
printf("after deletion the linked list be a\n");
display(head);
printf("\n");
printf("the number of nodes will be\n");
count(head);
insert(&head,333,20);
printf("\n");
display(head);
reverse(&head);
printf("\n");
printf("this is reverse the nodes\n");
display(head);
reversing(head);
printf("\n");
printf("this is reverse printing\n");
display(head);
printf("\n");
printf("the mid element is\n");
mid(head);

}
void addatbeg(struct node **q,int num)
{
struct node temp;
temp = (struct node)malloc(sizeof(struct node));
temp->data = num;
temp->next = q;
q = temp;
}
void display(struct node *q)
{
while(q!=NULL)
{
printf("%d ",q->data);
q = q->next;
}
}
void append(struct node**q,int num)
{
struct node temp,r;
if(q==NULL)
{
temp = (struct node)malloc(sizeof(struct node));
temp->data = num;
temp->next = q;
q = temp;
}
else
{
temp = q;
while(temp->next!=NULL)
temp = temp->next;
r = (struct node*)malloc(sizeof(struct node));
r->data = num;
r->next = NULL;
temp->next = r;
}
}
void delete(struct node **q,int num)
{
struct node* temp,old;
temp = q;
while(temp!=NULL)
{
if(temp->data == num)
{
if(temp == q)
{
q = NULL;
free(temp);
}
else
{
old->next = temp->next;
free(temp);
}
}
else
{

old = temp;
temp = temp->next;
}

}
}
void count(struct node*q)
{
int c=0;
while(q!=NULL)
{
c++;
q = q->next;
}
printf("the nuber of nodes = %d\n",c);

}
void insert(struct node**q,int num,int n)
{
struct node temp,r;
temp = q;
while(temp!=NULL)
{
if(temp->data == n)
{
r = (struct node)malloc(sizeof(struct node));
r->data = num;
r->next = temp->next;
temp->next = r;
}
temp = temp->next;
}
}
void reverse(struct node**q)
{
struct node temp,prev=NULL,rev;
temp = q;
while(temp!=NULL)
{
rev = temp;
temp = temp->next;
rev->next = prev;
prev = rev;
}
q = rev;
}
void reversing(struct nodeq)
{
if(q==NULL)
{
reversing(q->next);
printf("%d ",q->data);
}
}
void mid(struct node*q)
{
int m=0;
if(q->next==NULL)
{
printf("only one node is there\n");
}
if(q->next->next==NULL)
{
printf(" two nodes is there\n");
}
while( (q->next!=NULL) && (q->next->next!=NULL))
{
m++;
q = q->next;
}
printf("%d\n",m);
}

-------------------------------------------------------------------------------------------------------------------------------------------------

30 20 10
30 20 10 60 70
after deletion the linked list be a
30 20 10 70
the number of nodes will be
the nuber of nodes = 4

30 20 333 10 70
this is reverse the nodes
70 10 333 20 30
this is reverse printing
70 10 333 20 30
the mid element is
3