## Write a function that counts the number of times a given int occurs in a Linked List

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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