Converting a number to binary java code

Question

I am writing a code to convert a number to binary representation.Here's my code.It doesn't give the correct answer but I can't figure out where I am making the mistake.If someone could point out where my mistake and how to correct it I would be grateful.

public class ConvertBinary{
    public static String convertBinary(int n){
        String s="";
        while(n>0){
            s+=(n%2);

            n=n/2;
        }


        int len=s.length();
        String[] binary=s.split("");
        String[] binaryCopy=new String[s.length()];

        for(int i=0;i<len;i++){
            binaryCopy[i]=binary[len-i-1];
            s+=binaryCopy[i];
        }
        return s;

    }

    public static void main (String args[]){
        System.out.println(convertBinary(19));
    }
}

Answer 1

Apart from all these answers the problem with your code was that you are not clearing the string before reversing it. So before the for loop just put s = "" and you code should work fine.. :)

Based on comment

public class ConvertBinary {
    public static String convertBinary(int n) {
        String s = "";
        while (n > 0) {
            s += (n % 2);

            n = n / 2;
        }

        int len = s.length();
        String[] binary = s.split("");
        String[] binaryCopy = new String[s.length()];

        s = "";

        for (int i = 0; i < len; i++) {
            binaryCopy[i] = binary[len - i - 1];
            s += binaryCopy[i];
        }
        return s;

    }

    public static void main(String args[]) {
        int num = 4;
        System.out.println(convertBinary(num));
        System.out.println(Integer.toBinaryString(num));
    }
}

Answer 2

public static String convertBinary(int n){
        String s="";
        while(n>0){
            s+=(n%2);

            n=n/2;
        }



    return (new StringBuffer(s).reverse().toString());
}

Answer 3

If you're looking for error in your implementation, you'd rather put:

  s = (n % 2) + s;

isntead of

  s+=(n%2);

so the code'll be

  // n should be positive
  public static String convertBinary(int n){
    if (n == 0)
      return "0";

    String s = "";

    // for is more compact than while here
    for (; n > 0; n /= 2)
      s = (n % 2) + s;

    return s;
  }

however in real life

  Integer.toString(n, 2);

is much more convenient

Answer 4

Use Java standard library: http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#toString%28int,%20int%29

public class ConvertBinary{
    public static String convertBinary(int n){
        return Integer.toString(n, 2);
}

    public static void main (String args[]){
        System.out.println(ConveryBinary.convertBinary(19));
    }
}

EDIT: As @Holger says, there's also a toBinaryString: http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#toBinaryString%28int%29

public static String convertBinary(int n){
    return Integer.toBinaryString(n);
}