Technical Stuff

#include<stdio.h>
#define STACKSIZE 20
//stack size is 20
int stack[STACKSIZE]={0};
//top of the stack initially -1 which means empty
int tos=-1;
void push(int ele)
{
 if(tos>=STACKSIZE)
 {
 printf("err: Stack is full\n");
 }
 else
 {
 tos++;
 stack[tos] = ele; 
 }
}
int pop()
{
 if(tos==-1)
 {
 printf("err: stack is empty\n");
 }
 else
 {
 int temp = stack[tos];
 tos--;
 printf("poped element from the stack is %d\n",temp);
 return temp; 
 }
}
void display()
{
 int i;
 for(i=0;i<=tos;i++)
 printf(" %d\n",stack[i]);
 if(tos== -1)
 {
 printf("err: stack is empty\n");
 return;
 }
}
int main()
{
 int ch,ele;
 printf("enter the choice 1-push 2-pop 3-display 9-exit\n");
 scanf("%d",&ch);
 while(ch!=9)
 {
 switch(ch)
 {
 case 1: printf("Enter the ele to push\n");
 scanf("%d",&ele);
 push(ele);
 break;
 case 2: ele = pop();
 break;
 case 3: display();
 break;
 case 4:
 break;
 default:
 break;
 }
 printf("enter the choice 1-push 2-pop 3-display 9-exit\n");
 scanf("%d",&ch);
 }
}

No comments:

Labels: C program, Data Structures, Programming in C

Below is the C program for rearranging the linked list (not copying the elements), in such a way that, first odd elements next even elements should come as shown below. I faced this question in Amazon Interview first round. So Sharing the program here.

Input list: 1 2 3 4 5 6 7 8 9 10
OutputList: 1 3 5 7 9 2 4 6 8 10

#include<stdio.h>
#include<stdlib.h>
 
//linked list structure
struct node
{
 int info;
 struct node *next;
};
 
//making typdef so we can use Node instead of 'struct node'
typedef struct node Node;
//rearranging the linked list in such a way that odd no.s first 
//and then even no.
Node* oddEven(Node* h)
{
 Node* tempNode = h;
 Node* oddEnd = NULL,*oddHead = NULL;
 Node* evenHead = NULL, *evenEnd = NULL;
 while( tempNode != NULL)
 {
 if(!(tempNode->info%2))
 {
 if(evenHead == NULL)
 evenHead = tempNode;
 else
 evenEnd->next = tempNode;
 evenEnd = tempNode;
 }
 else
 {
 if(oddHead == NULL)
 oddHead = tempNode;
 else
 oddEnd->next = tempNode;
 oddEnd = tempNode;
 }
 tempNode = tempNode->next;
 }
 h = oddHead;
 oddEnd->next = evenHead;
 evenEnd->next = NULL;
 return h;
}
 
//inserting node or creating the list or adding the element @ end
Node* insert(Node *h, int info)
{
 Node *tempHead=h;
 Node *newNode = (Node*)malloc(sizeof(Node));
 newNode->info = info;
 newNode->next = NULL;
 if(tempHead == NULL) // if the list has zero elements. make new node as a head
 {
 h=newNode;
 }
 else if(tempHead->next==NULL) // if the list is having only one node
 {
 tempHead->next = newNode;
 }
 else
 {
 Node *tempNode = h;
 while(tempNode->next != NULL) // if the list has more than one node, so moving to the last node
 {
 tempNode = tempNode->next;
 }
 tempNode->next = newNode; // appending the new node at the end
 }
 return h;
}
 
/*****************************************************************************
for displaying the nodes in the list
*****************************************************************************/
void display(Node *h)
{
 Node *temp = h;
 while (temp->next!=NULL)
 {
 printf("%d->",temp->info);
 temp = temp->next;
 }
 printf("%d\n",temp->info);
}
 
int main()
{
 Node *head = NULL;
 int i,n,element,choice,pos,size;
 for (i=1;i<20;i++)
 {
 head = insert(head,i);
 }
 display(head);
 head=oddEven(head);
 display(head);
}

Output:

Input Linked List:
1->2->3->4->5->6->7->8->9->10->11->12->13->14->15->16->17->18->19

Resulted List:
1->3->5->7->9->11->13->15->17->19->2->4->6->8->10->12->14->16->18

Labels: Data Structures, Linked List

c ^ a = b




Before Operation:
After Operation:

After executing above statement, the current node moves to the next node as shown below.

Labels: Data Structures, Linked List