Prime Factor Constructor

Question

I am making a two classes, the constructing class and the main method where I read a number from a user input and spits out the prime factorizations of the number, code is with Java.

For Example:
Enter Number: 150
5
5
3
2

However, for my program I'm getting the entire list of factors.

Example:
Enter Number: 150
150
75
50
25
5
3
1

How would I change this up to get prime factors?

Main Method:

import java.util.*;

public class FactorPrinter
{
    public static void main(String[] args)
    {
        Scanner scan = new Scanner(System.in);
        System.out.print("Enter a integer: ");
        String input1 = scan.nextLine();
        int input = Integer.parseInt(input1);
        FactorGenerator factor = new FactorGenerator(input);
        System.out.print(factor.getNextFactor());

        while (!factor.hasMoreFactors())
        {
            System.out.print(factor.getNextFactor());
        }
     }  
}

Here is my class:

public class FactorGenerator 
{
    private int num;
    private int nextFactor;

    public FactorGenerator(int n)
    {
        num = nextFactor = n;
    }

    public int getNextFactor()
    {
        int i = nextFactor - 1 ;

        while ((num % i) != 0)
        {
            i--;
        }

        nextFactor = i;
        return i;
    }

    public boolean hasMoreFactors()
    {
        if (nextFactor == 1)
        {
            return false;
        }
        else
        {
            return true;
        }
    }
}

Answer 1

Following method will print prime factors for a given number. It contains few optimizations as well.

public static void primeFactors(int n) {
    if (n <= 1) {
        return;
    }
    while (n % 2 == 0) {
        System.out.println(2);
        n = n / 2;
    }
    while (n % 3 == 0) {
        System.out.println(3);
        n = n / 3;
    }
    for (int i = 5; i * i <= n; i += 6) {
        while (n % i == 0) {
            System.out.println(i);
            n = n / i;
        }
        while (n % (i + 2) == 0) {
            System.out.println(i + 2);
            n = n / (i + 2);
        }
    }
    if (n > 3) {
        System.out.println(n);
    }
}

Answer 2

A corrected version of @Bohemian's deleted answer:

for (int i = 2; input > 1 && i <= input; i++)
{
    if (input % i == 0)
    {
        System.out.print(i+" ");
        do
        {
            input /= i;
        } while (input % i == 0);
    }
}

There are much faster algorithms, e.g. Knuth The Art of Computer Programming, Vol I, #4.5.4 algorithm C, which derives from Fermat, but note that there is an important correction on his website. He gives a good testing value as 8616460799L as it has two rather large prime factors.

Answer 3

You can add a method to check if the factor being returned is a prime factor or not : Try something like this :

public bool isPrime(int number) {
    int i;
    for (i=2; i*i<=number; i++) {
        if (number % i == 0) return false;
    }
    return true;
}

And you can use this as :

 public class FactorPrinter
{
    public static void main(String[] args)
    {
        //your initial code

        while (!factor.hasMoreFactors())
        {
            int nextFactor= factor.getNextFactor()

            if(isPrime(nextFactor))
           {
            System.out.print();
           }
        }
     }  
}