This C Program find the number of occurrences of all elements in a linked list.
Here is source code of the C Program to find the number of occurrences of all elements in a linked list. The C program is successfully compiled and run on a Linux system. The program output is also shown below.
/*
* C Program to Find Number of Occurences of All Elements in a Linked List
*/
#include <stdio.h>
#include <stdlib.h>
struct node
{
int num;
struct node *next;
};
struct node_occur
{
int num;
int times;
struct node_occur *next;
};
void create(struct node **);
void occur(struct node *, struct node_occur **);
void release(struct node **);
void release_2(struct node_occur **);
void display(struct node *);
void disp_occur(struct node_occur *);
int main()
{
struct node *p = NULL;
struct node_occur *head = NULL;
int n;
printf("Enter data into the list\n");
create(&p);
printf("Displaying the occurence of each node in the list:\n");
display(p);
occur(p, &head);
disp_occur(head);
release(&p);
release_2(&head);
return 0;
}
void occur(struct node *head, struct node_occur **result)
{
struct node *p;
struct node_occur *temp, *prev;
p = head;
while (p != NULL)
{
temp = *result;
while (temp != NULL && temp->num != p->num)
{
prev = temp;
temp = temp->next;
}
if (temp == NULL)
{
temp = (struct node_occur *)malloc(sizeof(struct node_occur));
temp->num = p->num;
temp->times = 1;
temp->next = NULL;
if (*result != NULL)
{
prev->next = temp;
}
else
{
*result = temp;
}
}
else
{
temp->times += 1;
}
p = p->next;
}
}
void create(struct node **head)
{
int c, ch;
struct node *temp, *rear;
do
{
printf("Enter number: ");
scanf("%d", &c);
temp = (struct node *)malloc(sizeof(struct node));
temp->num = c;
temp->next = NULL;
if (*head == NULL)
{
*head = temp;
}
else
{
rear->next = temp;
}
rear = temp;
printf("Do you wish to continue [1/0]: ");
scanf("%d", &ch);
} while (ch != 0);
printf("\n");
}
void display(struct node *p)
{
while (p != NULL)
{
printf("%d\t", p->num);
p = p->next;
}
printf("\n");
}
void disp_occur(struct node_occur *p)
{
printf("***************************\n Number\tOccurence\n***************************\n");
while (p != NULL)
{
printf(" %d\t\t%d\n", p->num, p->times);
p = p->next;
}
}
void release(struct node **head)
{
struct node *temp = *head;
*head = (*head)->next;
while ((*head) != NULL)
{
free(temp);
temp = *head;
(*head) = (*head)->next;
}
}
void release_2(struct node_occur **head)
{
struct node_occur *temp = *head;
*head = (*head)->next;
while ((*head) != NULL)
{
free(temp);
temp = *head;
(*head) = (*head)->next;
}
}
$ cc occurence.c $ ./a.out Enter data into the list Enter number: 1 Do you wish to continue [1/0]: 1 Enter number: 2 Do you wish to continue [1/0]: 1 Enter number: 3 Do you wish to continue [1/0]: 1 Enter number: 2 Do you wish to continue [1/0]: 1 Enter number: 4 Do you wish to continue [1/0]: 1 Enter number: 2 Do you wish to continue [1/0]: 1 Enter number: 6 Do you wish to continue [1/0]: 1 Enter number: 1 Do you wish to continue [1/0]: 0 Displaying the occurence of each node in the list: 1 2 3 2 4 2 6 1 *************************** Number Occurence *************************** 1 2 2 3 3 1 4 1 6 1
Sanfoundry Global Education & Learning Series – 1000 C Programs.
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