Binary format of an integer

Question

I have a method that prints out binary formar of numbers:

private static void binary(int n){
   for(int i = 1; i < Math.pow(2, n); i++)
      System.out.println(Integer.toBinaryString(i));
   }
} 

Output is like for n = 3:

1
10
11
100
101
110
111

Is there a way to print out like this:

001
010
011
100
101
110
111

Answer 1

    private static void binary(int n){
       String t = ""; int N = 1<<n;
       for (int i=0; i<n;i++) t += "0";
       for (int i = 1; i < N; i++) {
             String s = Integer.toBinaryString(i);
             System.out.println(t.substring(s.length())+s);
       }
    } 

Answer 2

I would write it as

private static void binary(int n){
   for(long i = 0, max = 1 << n; i < max; i++) {
      String s = Long.toBinaryString(i);
      while (s.length() < n) s = '0' + s;
      System.out.println(s);
   }
} 

or more efficiently

private static void binary(int n) {
    char[] chars = new char[n];
    for (long i = 0, max = 1 << n; i < max; i++) {
        for (int j = 0; j < n; j++)
            chars[j] = (char) (((i >>> (n - j - 1)) & 1) + '0');
        System.out.println(chars);
    }
}

binary(3);

prints

000
001
010
011
100
101
110
111

Answer 3

Quick/Dirty version:

System.out.println(Integer.toBinaryString(i+(1<<n)).substring(1));

(obvious flaw if n == the number of bits in an integer, or if you have more bits in your integer than you want to print... so limited, but maybe a bit simpler if you know your inputs)

Answer 4

Try:

String.format("%" + n + "s", Integer.toBinaryString(i)).replace(' ', '0')

It may not be the most efficient solution, but it's simple and short enough.

Answer 5

private static void binary(int n){
   for(int i = 1; i < Math.pow(2, n); i++)
      System.out.println(String.format("%04d", Integer.parseInt(Integer.toBinaryString(i))));
   }
} 

How can I pad an integers with zeros on the left?