Link List Count Function

NIELIT O Level Solved Paper

January 2016

Q 9 B. Define the structure of a node of a singly linked list and use it to write a function to count the
number of nodes in a singly linked list. The function should accept a pointer to the first node of
the list and it should return the number of nodes in the list.

Solution : 

/*   ************************************************

           Girfa : Student Help

           Counting node in link list

           for more program visit : http://girfahelp.blogspot.com/p/c-language-assignment.html

     *************************************************/

typedef struct n

{

 int data;

 struct n *next;

}node;

/* Function Prototype */

node *start=0; /* Global Variable point first node*/

int count();

void add(int);

void disp();

#include<stdio.h>

#include<conio.h>

void main()

{

     int opt,n;

     do

     {

           clrscr();

           printf("\n1. Add\n2. Count\n3. Display\n0. Exit\nEnter your choice>> ");

           scanf("%d",&opt);

           switch(opt)

           {

                case 1:

                printf("Enter  number to add in list>> ");

                scanf("%d",&n);

                add(n);

                break;

                case 2:

                     printf("\nTotal %d node in link list>> ",count(start));

                     getch();

                     break;

                case 3:

                     disp();

                     getch();

                case 0:

                     break;

                default:

                     printf("Invalid Choice");

                     getch();

           }

       }while(opt!=0);

}

int count()

{

     node *pt;

     int i=0;

     for(pt=start;pt!=NULL;pt=pt->next)

           i++;

     return i;

}

void add(int n)

{

           node *nw=(node*) malloc(sizeof(node));

           nw->next=NULL;

           nw->data=n;

           if(start==NULL)

                start=nw;

           else

           {

                nw->next=start;

                start=nw;

           }

}

void disp()

{

     node *pt;

     for(pt=start;pt!='\0';pt=pt->next)

           printf("[%d]",pt->data);

}

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