Java - Roman Numeral validity

Question

I am writing a program to add two roman numerals without converting to any bases. I have everything working except I am not sure how to check if my input String is a valid roman numeral or not.

These are the rules to check validity:

Five in a row of any digit is not allowed
Some digits are allowed in runs of up to 4. They are I,X,C, and M. The others (V,L,D) can only appear singly.
Some lower digits can come before a higher digit, but only if they appear singly. E.g. "IX" is ok but "IIIX" is not.
But this is only for pairs of digits. Three ascending numbers in a row is invalid. E.g. "IX" is ok but "IXC" is not.
A single digit with no runs is always allowed

I have not really been able to make much progress with this step and don't have anything working yet. Any help would be great!

What code do you have so far? What does it do? What did you expect it to do? — Robert, Jul 06 '16 at 03:35
Number 2 premise if FALSE: http://www.roman-numerals.org/chart100.html — Grayson, Jul 06 '16 at 17:08

score 9 · Answer 1 · edited May 23 '17 at 11:46

9

Why not use regular expression:

boolean valid = word.matches("^M{0,4}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})$");

Look at the post of paxdiablo: How do you match only valid roman numerals with a regular expression?

edited May 23 '17 at 11:46

Community

1
1

answered Mar 28 '17 at 01:42

Jeff_Alieffson

2,672
29
34

score 1 · Answer 2 · answered Jul 06 '16 at 01:48

1

Loop through each character in the string. Use a if conditions to check which character it is. Use the if conditions to check for the roman numeral rule violations in the adjacent characters to the selected character.

for(int i = 0; i < s.length(); i++){  
    if (s[i] == 'V'){    
        **Check if the character before of after is also 'V'. Then it is a violation  
    }     
    else if(s[i] == 'L'){  
        **Conditions for 'L' etc.  
    }  
}

answered Jul 06 '16 at 01:48

Ivantha

491
2
9
27

It seems like thats simplifying that too much isn't it? I'm trying to code that and there are wayy too many conditions to account for in that case. – RajveerParikh Jul 06 '16 at 03:25

score 1 · Answer 3 · answered Jul 06 '16 at 17:05

This is what I have come up with based on my rules. Any thoughts on refactoring this to make it simpler?

public static boolean checkValidity (String s1, HashSet<Character> romanNumerals){
    HashSet<Character> alreadyContained = new HashSet<Character>();
    if (s1.length() == 1 && romanNumerals.contains(s1.charAt(0))){
        return true;
    }
    int i = 0;
    while (i < s1.length()){
        if (s1.charAt(i) == 'M'){
            if (alreadyContained.contains('M')){
                return false;
            }
            int count = 1;
            i++;
            while (s1.charAt(i) == 'M'){
                i++;
                count++;
            }
            alreadyContained.add('M');
            if (count >= 5){
                return false;
            }
        }
        else if (s1.charAt(i) == 'D'){
            if (alreadyContained.contains('D')){
                return false;
            }
            alreadyContained.add('D');
            if (!alreadyContained.contains('M')){
                alreadyContained.add('M');
            }
            i++;
            if ((i < s1.length()) && (s1.charAt(i) == 'D')){
                return false;
            }

        }
        else if (s1.charAt(i) == 'L'){
            if (alreadyContained.contains('L')){
                return false;
            }
            alreadyContained.add('L');
            if (!alreadyContained.contains('M')){
                alreadyContained.add('M');
            }
            if (!alreadyContained.contains('D')){
                alreadyContained.add('D');
            }
            if (!alreadyContained.contains('C')){
                alreadyContained.add('C');
            }
            i++;
            if ((i < s1.length()) && (s1.charAt(i) == 'L')){
                return false;
            }
        }
        else if (s1.charAt(i) == 'V'){
            if (alreadyContained.contains('V')){
                return false;
            }
            alreadyContained.add('V');
            if (!alreadyContained.contains('M')){
                alreadyContained.add('M');
            }
            if (!alreadyContained.contains('D')){
                alreadyContained.add('D');
            }
            if (!alreadyContained.contains('C')){
                alreadyContained.add('C');
            }
            if (!alreadyContained.contains('L')){
                alreadyContained.add('L');
            }
            if (!alreadyContained.contains('X')){
                alreadyContained.add('X');
            }
            i++;
            if ((i < s1.length()) && (s1.charAt(i) == 'V')){
                return false;
            }

        }
        else if (s1.charAt(i) == 'C'){
            if (alreadyContained.contains('C')){
                return false;
            }
            int count = 1;
            i++;
            if ((i < s1.length()) &&(s1.charAt(i) == 'M' || s1.charAt(i) == 'D')){
                i++;
            }
            else if (i < s1.length() && s1.charAt(i) == 'C'){
                while ((i < s1.length()) && (s1.charAt(i) == 'C')){
                    i++;
                    count++;
                }
            }
            alreadyContained.add('C');
            if (!alreadyContained.contains('M')){
                alreadyContained.add('M');
            }
            if (!alreadyContained.add('D')){
                alreadyContained.add('D');
            }
            if (count >= 5){
                return false;
            }
        }
        else if (s1.charAt(i) == 'X'){
            if (alreadyContained.contains('X')){
                return false;
            }
            int count = 1;
            i++;
            if ((i < s1.length()) && (s1.charAt(i) == 'D' || s1.charAt(i) == 'M')){
                return false;
            }
            while ((i < s1.length()) && s1.charAt(i) == 'X'){
                i++;
                count++;
            }
            alreadyContained.add('X');
            if (count >= 5){
                return false;
            }
        }
        else if (s1.charAt(i) == 'I'){
            if (alreadyContained.contains('I')){
                return false;
            }
            alreadyContained.add('I');
            i++;
            int count = 1;
            if ((i < s1.length()) && (s1.charAt(i) != 'I' && s1.charAt(i) != 'X' && s1.charAt(i) != 'V')){
                return false;
            }
            else if (i < s1.length() && s1.charAt(i) == 'I'){
                while (i < s1.length() && s1.charAt(i) == 'I'){
                    i++;
                    count++;
                }
                if (count >= 4){
                    return false;
                }
            }
        }
        else if (!romanNumerals.contains(s1.charAt(i))){
            return false;
        }
    }
    return true;
}

Java - Roman Numeral validity

3 Answers3