Roman Numeral to Integer

Question

i am trying to find the simplest algorithm to covert a Roman Numeral to int.I have this code from Rosetta code but there are few things in the code that doesnt make sense.

In the for loop in the code below how is the roman numeral being iterated through? Can any explain how the nested for loop here work or do you have a simpler way of writing this algorithm

public class Roman {

    enum Numeral {
        I(1), IV(4), V(5), IX(9), X(10), XL(40), L(50), XC(90), C(100), CD(400), D(500), CM(900), M(1000);
        int weight;

        Numeral(int weight) {
            this.weight = weight;
        }
    };

    public static String roman(long n) {

        if( n <= 0) {
            throw new IllegalArgumentException();
        }

        StringBuilder buf = new StringBuilder();

        final Numeral[] values = Numeral.values();
        for (int i = values.length - 1; i >= 0; i--) {
            while (n >= values[i].weight) {
                buf.append(values[i]);
                n -= values[i].weight;
            }
        }
        return buf.toString();
    }

    public static void test(long n) {
        System.out.println(n + " = " + roman(n));
    }

    public static void main(String[] args) {
        test(1999);
        test(25);
        test(944);
        test(0);
    }

Unless you phrased your question backwards, this enum is for converting Integers to Roman Numerals, not Roman Numerals to Integers. — ian.shaun.thomas, Jan 21 '16 at 19:21
Possible duplicate of [Converting Roman Numerals To Decimal](http://stackoverflow.com/questions/9073150/converting-roman-numerals-to-decimal) — ian.shaun.thomas, Jan 21 '16 at 19:23

score 0 · Answer 1 · answered Apr 02 '15 at 18:05

The roman literals are being iterated from highest to lowest. This makes perfect sense. This is how one should do it because you want to extract the biggest item first. Once you're done with it, you want to extract the second biggest item and so on. While extracting an item it may fit 0,1,2,3,... times into the number you currently have left. That's the meaning/purpose of this loop there.

        while (n >= values[i].weight) {
            buf.append(values[i]);
            n -= values[i].weight;
        }

And I think that's pretty much the simplest way to write this algorithm. You just need to grasp the idea.

It is referring to the value of each Roman literal/numeral e.g. for IV the weight is 4. — peter.petrov, Apr 03 '15 at 14:16

score 0 · Answer 2 · answered Nov 14 '16 at 22:01

public int romanToInt(String s) {
        int total = 0, currentVal = 0, prevVal = 0;

        for(int i=s.length()-1; i>=0; i--) {
            switch(s.charAt(i)) {
                case 'I' : currentVal = 1; break;
                case 'V' : currentVal = 5; break;
                case 'X' : currentVal = 10; break;
                case 'L' : currentVal = 50; break; 
                case 'C' : currentVal = 100; break;
                case 'D' : currentVal = 500; break;
                case 'M' : currentVal = 1000; break; 
                default: break;
            }
            total += (currentVal < prevVal) ? -1 * currentVal : currentVal;
            prevVal = currentVal;
        }

        return total;
    }

Roman Numeral to Integer

2 Answers2