FizzBuzz problem | Java program

Write a program that outputs the string representation of numbers from 1 to n. but for multiples of three it should output "Fizz" instead of the number and for the multiples of five output "Buzz". For numbers which are multiples of both three and five output "FizzBuzz".

ALGORITHM

JAVAfizzbuzz(n):
            list=[];
            for i from 1 to n:
                if i%5==0 && i%3==0:
                    list.add("FizzBuzz");
                else if i%3==0:
                    list.add("Fizz");
                else if i%5==0:
                    list.add("Buzz");
                else:
                    list.add(Integer.toString(i));
            return list;

JAVA
package bracecoder;

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class FizzBuzz {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Scanner scan=new Scanner(System.in);
		System.out.println("Enter a Number : ");
		int n=scan.nextInt();
		List res = fizzBuzz(n);
		System.out.println(res);
	}

	private static List fizzBuzz(int n) {
		// TODO Auto-generated method stub
            List res = new ArrayList();
            for (int i = 1; i <= n; i++) {
                if (i % 3 == 0 && i % 5 == 0) {
                    res.add("FizzBuzz");
                } else if (i % 3 == 0) {
                    res.add("Fizz");
                } else if (i % 5 == 0) {
                    res.add("Buzz");
                } else {
                    res.add(Integer.toString(i));
                }
            }
            return res;
	}
}

OUTPUT

ALGORITHM

JAVA      fizzbuzz(n):
            list=[];
            i3=0;i5=0;
            for i from 1 to n:
                i3=++;i5++;
                if i5==5 && i3==3:
                    list.add("FizzBuzz"); i3=0;i5=0;
                else if i3==3:
                    list.add("Fizz"); i3=0;
                else if i5==5:
                    list.add("Buzz"); i5=0;
                else:
                    list.add(Integer.toString(i));
            return list;

JAVApackage bracecoder;

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class FizzBuzz {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Scanner scan=new Scanner(System.in);
		System.out.println("Enter a Number : ");
		int n=scan.nextInt();
		List res = fizzBuzz(n);
		System.out.println(res);
		

	}

	private static List fizzBuzz(int n) {
		// TODO Auto-generated method stub
		List res = new ArrayList();
		int i3=0, i5=0;
        for (int i = 1; i <= n; i++) {
        	i3++;
        	i5++;
            if (i3==3 && i5==5) {
                res.add("FizzBuzz");
                i3=0;
                i5=0;
            } else if (i3==3) {
                res.add("Fizz");                
                i3=0;
            } else if (i5==5) {
                res.add("Buzz");
                i5=0;
            } else {
                res.add(Integer.toString(i));
            }
        }
        return res;
	}

}

OUTPUT

Frequently asked questions

Why must the "FizzBuzz" check come first?

Numbers like 15 are multiples of both 3 and 5. If the multiple-of-3 check ran first, 15 would print "Fizz" and never reach "FizzBuzz".

How does the method without the % operator work?

It keeps two counters that go up by one for each number. When the 3-counter reaches 3 it prints "Fizz" and resets, when the 5-counter reaches 5 it prints "Buzz" and resets, and when both reach their limit together it prints "FizzBuzz".

What is the time complexity?

Both methods visit each number from 1 to n once, so they run in O(n) time and store n strings in the result list.