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

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.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the 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:

## C/C++

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

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 */

/* move the head to point to the new node */
}

/* Counts the no. of occurences of a node
int count(struct Node* head, int search_for)
{
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()
{

/* Use push() to construct below list
1->2->1->3->1  */

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

## Java

```// Java program to count occurrences in a linked list
{

class Node
{
int data;
Node next;
Node(int d) {data = d; next = null; }
}

/* Inserts a new Node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);

/* 3. Make next of new Node as head */

/* 4. Move the head to point to new Node */
}

/* Counts the no. of occurences of a node
int count(int search_for)
{
int count = 0;
while (current != null)
{
if (current.data == search_for)
count++;
current = current.next;
}
return count;
}

/* Drier function to test the above methods */
public static void main(String args[])
{

/* Use push() to construct below list
1->2->1->3->1  */
llist.push(1);
llist.push(2);
llist.push(1);
llist.push(3);
llist.push(1);

/*Checking count function*/
System.out.println("Count of 1 is "+llist.count(1));
}
}
// This code is contributed by Rajat Mishra
```

## Python

```
# Python program to count the number of time a given
# int occurs in a linked list

# Node class
class Node:

# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None

def __init__(self):

# Counts the no . of occurances of a node
def count(self, search_for):
count = 0
while(current is not None):
if current.data == search_for:
count += 1
current = current.next
return count

# Function to insert a new node at the beginning
def push(self, new_data):
new_node = Node(new_data)

def printList(self):
while(temp):
print temp.data,
temp = temp.next

# Driver program
llist.push(1)
llist.push(3)
llist.push(1)
llist.push(2)
llist.push(1)

# Check for the count function
print "count of 1 is %d" %(llist.count(1))

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

```

Output:
`count of 1 is 3`

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

# GATE CS Corner    Company Wise Coding Practice

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