Here is a solution.

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. 

Implementation:

#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));    
    getchar();
}

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

         

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