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Write a program to create a single link list and display the link list.
1 Answer
written 7.6 years ago by |
# include <stdio.h>
# include <stdlib.h>
struct node
{
int data;
struct node *link;
};
struct node *insert(struct node *p, int n)
{
struct node *temp;
/* if the existing list is empty then insert a new node as the
starting node */
if(p==NULL)
{
p=(struct node *)malloc(sizeof(struct node)); /* creates new node
data value passes
as parameter */
if(p==NULL)
{
printf("Error\n");
exit(0);
}
p-> data = n;
p-> link = p; /* makes the pointer pointing to itself because it
is a circular list*/
}
else
{
temp = p;
/* traverses the existing list to get the pointer to the last node of
it */
while (temp-> link != p)
temp = temp-> link;
temp-> link = (struct node *)malloc(sizeof(struct node)); /*
creates new node using
data value passes as
parameter and puts its
address in the link field
of last node of the
existing list*/
if(temp -> link == NULL)
{
printf("Error\n");
exit(0);
}
temp = temp-> link;
temp-> data = n;
temp-> link = p;
}
return (p);
}
void printlist ( struct node *p )
{
struct node *temp;
temp = p;
printf("The data values in the list are\n");
if(p!= NULL)
{
do
{
printf("%d\t",temp->data);
temp=temp->link;
} while (temp!= p);
}
else
printf("The list is empty\n");
}
void main()
{
int n;
int x;
struct node *start = NULL ;
printf("Enter the nodes to be created \n");
scanf("%d",&n);
while ( n -- > 0 )
{
printf( "Enter the data values to be placed in a node\n");
scanf("%d",&x);
start = insert ( start, x );
}
printf("The created list is\n");
printlist ( start );
}