How do I format a number left-padded without using String.format in Java 6?

Question

For example, I need to grab an unknown number, let's say 3, and find the binary (2^3) - 1 times, from 0 to 111 (0-7). Obviously, the number of digits I need depends on whatever number 'n' in 2^n.

So, if the number is 3, I would need the output to be:

Now obviously I can do this manually with a String.format("%03d", NumberInBinary) operation, but that's hardcoding it for 3 digits. I need to do the equivalent code with an unknown number of digits, how can I do that? (as in String.format("%0nd", yournumber) where n is the number of digits.)

probably a duplicate of stackoverflow.com/questions/625838/… — Matthew
– Matthew, Commented Mar 4, 2011 at 23:51

Bala R · Accepted Answer · 2011-03-04 23:55:23Z

5

if n = 4, NumberInBinary = 101;

String.format("%0"+n+"d", NumberInBinary);

with output

edited Mar 4, 2011 at 23:55

answered Mar 4, 2011 at 23:29

Bala R

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5 Comments

Matthew Over a year ago

this doesn't cover the enumeration and binary conversion aspect.

Matthew Over a year ago

also, I don't see what NumberInBinary would be for this to work.

Bala R Over a year ago

@Matthew read the question completly first. n and NumberInBinary are already available for the OP. he just needs to print it.

Bala R Over a year ago

@Matthew just output binary value, not find.

BlinkerBoy Over a year ago

StackOverflowException is correct, I have already found the binary number. And his solution worked perfectly, I had no idea you could do that, and it makes perfect sense since the inside of the quotes are treated as a String, so doing "%0"+n+"d" did the trick perfectly, adapting itself to every case. Thanks, brother. And thanks Matthew, you also helped me learn of an alternate method of finding the binary value. I just used a for loop, starting at int i = 0, going up until 2^n - 1, getting 'i', using Integer.toString(i, 2), parsing it, and outputting it using StackOverflowExceptions's method.

Costi Ciudatu · Accepted Answer · 2011-03-05 00:14:06Z

1

Why not make use of the already built-in Integer.toBinaryString() and just manually add the zeros using a StringBuilder ?

public static void main(String[] args) {
    int max = 5;
    for (int i = 0; i < Integer.MAX_VALUE; i++) {
        String binary = Integer.toBinaryString(i);
        if (binary.length() > max) {
            break;
        }
        System.out.println( prefixWithZeros(binary, max) );
    }
}

static String prefixWithZeros(String binary, int n) {
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < n - binary.length(); i++) {
        sb.append('0');
    }
    return sb.append(binary).toString();
}

answered Mar 5, 2011 at 0:14

Costi Ciudatu

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Comments

Matthew · Accepted Answer · 2011-03-04 23:34:15Z

You could use recursion:

public static void enumerate(String prefix, int remaining) {
    if (remaining == 0) {
        System.out.println(prefix);
    } else {
        enumerate(prefix + "0", remaining - 1);
        enumerate(prefix + "1", remaining - 1);
    }
}

and then call enumerate("", numberOfDigits);

faster, using StringBuffer:

public static void enumerate(StringBuffer prefix, int remaining) {
    if (remaining == 0) {
        System.out.println(prefix.toString());
    } else {
        enumerate(prefix.append('0'), remaining - 1);
        prefix.deleteCharAt(prefix.length() - 1);
        enumerate(prefix.append('1'), remaining - 1);
        prefix.deleteCharAt(prefix.length() - 1);
    }
}

and then call enumerate(new StringBuffer(), numberOfDigits);

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