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Tuesday, 11 July 2017 12:33

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

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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/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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Anish Sir

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