Given a singly linked list and a key, count number of occurrences of given key in linked list. For example, if given linked list is 1->2->1->2->1->3->1 and given key is 1, then output should be 4.

Algorithm:

1. Initialize count as zero.
2. Loop through each element of linked list:
     a) If element data is equal to the passed number then
        increment the count.
3. Return count. 

C Programming:

// C/C++ program to count occurrences in a linked list
#include<stdio.h>
#include<stdlib.h>

/* Link list node */
struct node
{
int data;
struct node* next;
};

/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));

/* put in the data */
new_node->data = new_data;

/* link the old list off the new node */
new_node->next = (*head_ref);

/* move the head to point to the new node */
(*head_ref) = new_node;
}

/* Counts the no. of occurences of a node
(search_for) in a linked list (head)*/
int count(struct node* head, int search_for)
{
struct node* current = head;
int count = 0;
while (current != NULL)
{
if (current->data == search_for)
count++;
current = current->next;
}
return count;
}

/* Drier program to test count function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;

/* Use push() to construct below list
1->2->1->3->1 */
push(&head, 1);
push(&head, 3);
push(&head, 1);
push(&head, 2);
push(&head, 1);

/* Check the count function */
printf("count of 1 is %d", count(head, 1));
return 0;
}

Output:

count of 1 is 3

Time Complexity: O(n)
Auxiliary Space: O(1)

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