Java program to check odd or even using recursion
Java program to check odd or even using recursion
In this tutorial, we discuss a concept of Java program to check odd or even using recursion
How do we identify whether a given number is odd or even?
when you divide a number by two and if the balance is zero, it is an even number
when you divided a number by two and if the balance is one, it is an odd number
Example of even number 2,4,6,8,…..
Example of odd number 1,3,5,7,…..
Here we will use a modular operator to display odd or even number in the given range.
if n%2==0, n is a even number
if n%2==1, n is a odd number
Here’s a Java program to check whether given number is odd or even using recursion
Program 1
// Java code to check the given number odd or even import java.util.Scanner; class OddEvenCheck{ //recursive function to check if a number is even public static Boolean isEven(int n){ //Base cases if(n==0){ return true;//0 is even }else if(n==1){ return false;//1 is odd }else{ return isEven(n-2); } } public static void main(String[] args) { Scanner input=new Scanner(System.in); System.out.println("Enter a number to check odd or even:"); int num=input.nextInt(); //Check if the number is even if(isEven(num)){ System.out.println(num+ " is even"); }else{ System.out.println(num+ " is odd"); } input.close(); } }
When the above code is executed, it produces the following result
Case 1
Enter a number to check odd or even:
45
45 is odd
Case 2
Enter a number to check odd or even:
90
90 is even
Explanation
The recursive function isEven(int n) checks whether the number n is even or odd recursively by subtracting 2 from the number until it reaches either 0 (even) or 1 (odd)
In the main method , the user is prompt to input a number , and based on the result from isEven(), the program gives the output whether are number is odd or even
This program works continually reducing the number by 2 and using the fact that even numbers reach 0, while odd numbers reach 1
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