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.
Java Programming:
// Java program to count occurrences in a linked list
class LinkedList
{
Node head; // head of list
/* Linked list Node*/
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 */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* Counts the no. of occurences of a node
(search_for) in a linked list (head)*/
int count(int search_for)
{
Node current = head;
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[])
{
LinkedList llist = new LinkedList();
/* 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
Output:
count of 1 is 3
Time Complexity: O(n)
Auxiliary Space: O(1)
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