- Given a single linked list, count number of times given element exist in a single linked list.
- e.g. if we want to count node = 10 exist in a single linked list (Fig 1), then output should be 2.
- Similarly, if we would like to find node = 20 in a single linked list, then output should be 3.
Algorithm – find number of times element exists in a single linked list
- Find number of times element inputData = 10 exists in a single linked list
- Declare variable count = 0
- defines number of time element exists in a single linked list
- Start traversing the linked list from head
- Compare node data with 10
- if data is equal
- increment the count ( count++)
- if data is equal
- Keeps on iterating till, we reach end of single linked list
- Compare node data with 10
- By the end traversal, we will get number of times element exists in a single linked list.
Time complexity of algorithm is O(n).
Program: number of times given element exists in a single linked list (java)
1.) CountElement Class:
- countElement metehod, counts the number of times element exists in a single linked list
- insert function is used to insert element to a single linked list
- insert function is used to create linked list
- print function is used to print the single linked list (head to tail)
package org.learn.Question; import org.learn.List.Node; public class CountElement { public static int countElement(Node head, int search) { int elementCount = 0; // 1 -> 2 -> 3 while (head != null) { if (head.data == search) { elementCount++; } head = head.next; } return elementCount; } public static void insert(Node head, int data) { while (head.next != null) head = head.next; head.next = new Node(data); } public static void print(Node head) { while (head != null) { System.out.printf("%d ", head.data); head = head.next; } System.out.println(""); } }
2.) App Class:
- We are creating the single linked list in main method.
- We are printing the single linked list using CountElement.print method.
- We are calling countElement method, to count number of times, element(s) exists in a single linked list.
package org.learn.Client; import org.learn.List.Node; import org.learn.Question.CountElement; public class App { public static void main(String[] args) { int[] data = { 10, 20, 10, 20, 50, 20, 50 }; Node head = new Node(data[0]); for (int count = 1; count < data.length; count++) CountElement.insert(head, data[count]); System.out.printf("Linked list is : "); CountElement.print(head); int search = 10; int count = CountElement.countElement(head, search); System.out.printf("Element %d exists %d times in linked list", search, count); search = 20; count = CountElement.countElement(head, search); System.out.printf("\nElement %d exists %d times in linked list", search, count); search = 50; count = CountElement.countElement(head, search); System.out.printf("\nElement %d exists %d times in linked list", search, count); } }
3.) Node Class:
- Node class representing the nodes of single linked list
package org.learn.List; public class Node { public int data; public Node next; public Node(int num) { this.data = num; this.next = null; } }
Output – find number of times given node exists in a linked list using java
Linked list is : 10 20 10 20 50 20 50 Element 10 exists 2 times in linked list Element 20 exists 3 times in linked list Element 50 exists 2 times in linked list
Download code – count given node/ element in single linked list in java