[Easy/Tough] Roman Numerals

[Jignesh Malavia]

Write a program that converts Roman numerals to Arabic and reverse.
Values:
I = 1 (one)
V = 5 (five)
X = 10 (ten)
L = 50 (fifty)
C = 100 (one hundred)
D = 500 (five hundred)
M = 1000 (one thousand)
Examples:
30 is written as XXX but 15 is not written as VVV
40 is written as XL but 45 is not written as VL
9 is written as IX but 49 is not written as IL and 99 is not written as IC
243 is written as CCXLIII
1510 is written as MDX
1999 is written as MCMXCIX
4,9,40,90,400,900 == IV,IX,XL,XC,CC,CM respectively
Rules:
1) No googling for any more rules or hints.
PS:
If the above is just plain boring, then you can solve this quiz or read about the 1999 problem

[Michael Dunn]

Here's my lash at it.

class RomanNumerals
{
  private static char numerals[] = new char[1001];
  public static void main(String args[])
  {
    numerals[1]='I';numerals[5]='V';numerals[10]='X';numerals[50]='L';
    numerals[100]='C';numerals[500]='D';numerals[1000]='M';
    String converted = "";
    try
    {
      int number = Integer.parseInt(args[0]);
      converted = toRomanNumerals(number);
      System.out.println(converted);
      System.out.println(toNumbers(converted));
    }
    catch(Exception e)
    {
      System.out.println("error - invalid number characters");
    }
    System.exit(0);
  }
  private static String toRomanNumerals(int number)
  {
    String conversion = "";
    if(number < 1 || number >= 4000) return "outside number range";
    else
    {
      for(int i = 3; i >= 0; i--)
      {
        int powNum = (int)Math.pow(10,i);
        int bit = number/powNum;
        number -= bit*powNum;
        if(bit==9) conversion += ""+numerals[powNum]+numerals[powNum*10];
        else if(bit == 4) conversion += ""+numerals[powNum]+numerals[powNum*10/2];
        else
        {
          if(bit >= 5)
          {
            conversion += numerals[powNum*10/2];
            bit -= 5;
          }
          for(int ii=0; ii < bit; ii++)
          {
            conversion += numerals[powNum];
          }
        }
      }
    }
    return conversion;
  }
  private static int toNumbers(String romanNumerals)
  {
    int x = 0, y = 0, number = 0;
    for(int i = 0; i < romanNumerals.length() - 1; i++)
    {
      for(int ii = 0; ii < numerals.length; ii++)
      {
        if(romanNumerals.charAt(i) == numerals[ii])
        {
          x = ii;
        }
        if(romanNumerals.charAt(i+1) == numerals[ii])
        {
          y = ii;
        }
      }
      if(x < y) number -= x;
      else number += x;
    }
    for(int i = 0; i < numerals.length; i++)
    {
      if(romanNumerals.charAt(romanNumerals.length()-1) == numerals[i])
      {
        number += i;
        break;
      }
    }
    return number;
  }
}

[Joe Pluta]

Here's my shot at it:

public class Roman {
    private static final int[] a2r1 =
    { 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };
    private static final String[] a2r2 =
    { "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" };
    
    private static final String r2a1 = "MDCLXVI";
    private static final int[] r2a2 = { 1000, 500, 100, 50, 10, 5, 1 };
    
public static void main(java.lang.String[] args) {
    System.out.println(toRoman(1999));
    System.out.println(toRoman(256));
    System.out.println(toRoman(44));
    System.out.println(toArabic("MCMXLI"));
    System.out.println(toArabic("MDLI"));
    System.out.println(toArabic("XVI"));
}
public static int toArabic(String roman) {
    int arabic = 0;
    int last = 0;
    for (int i = 0; i < roman.length(); i++)
    {
        int current = r2a2[r2a1.indexOf(roman.charAt(i))];
        if (current > last) arabic -= (last * 2);
        arabic += current;
        last = current;
    }
    return arabic;
}
public static String toRoman(int arabic) {
    StringBuffer sb = new StringBuffer();
    for (int i = 0; i < a2r1.length; i++)
    {
        while (arabic >= a2r1[i])
        {
            arabic -= a2r1[i];
            sb.append(a2r2[i]);
        }
    }
    return sb.toString();
}
}

This will throw an exception if the Roman numeral is incorrect; it doesn't do any nice error checking like Michael's, but it's short and simple.
Joe

[Jignesh Malavia]

Yup!
You both get an A+ :-)