如何将 Integer 转换为没有 Integer.toString Function 的字符串 [暂停]

Question

This is an extension to a previous question of mine, but I'm required to do this problem without this function. I have to use recursion only. Any thoughts?

An example; 20A would be:

2 x 12^2 + 0 x 12^1 + 10 x 12^0 = 2 x 144 + 0 x 12 + 10 x 1 = 288 + 0 + 10 = 298

public class Duodecimal {
      public static String toBase12(int n) {
        if (n < 10)
            return Integer.toString(n);
        if (n == 10)
            return "A";
        if (n == 11)
            return "B";
        return toBase12(n / 12) + toBase12(n % 12);
      }
    }

Answer 1

Well, you've figured it out for n == 10. Surely you can use the same technique for n == 9. If you want an alternate idea, imagine you have the string '0123456789AB'. String has a method to ask for the character at a given index. NOw, what digit is at position '9' in that string? What is at 10? Exactly. This method is charAt .

I bet you can finish this homework with this hint:)

Answer 2

With a little thought, you could extend this to convert any base from 2 thru 16.

1. Add a second parameter to the method and call it base or radix
2. Define a final String of digits as "0123456789ABCDEF"
3. Put in appropriate check(s) to ensure the radix is within an allowable range.