LeetCode - Roman to Integer

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

http://oj.leetcode.com/problems/roman-to-integer/

Solution:

First thing is to have function that convert roman char to int. They are I->1, V->5, X->10, L->50, C->100, D->500, M->1000.

And there are two rules in writing roman numbers. First is to simply add the char. Another is subtraction like 4 in roman number is IV which means V-I.

So let's see what happen in 1~10, which are I, II, III, IV, V, VI, VIII, IX and X. It seems we find some laws in it.

If s[i] > s[i-1], we can simply add the numer, else we need to add (s[i] - s[i-1]). And since we have already added s[i-1] before, we need to subtract s[i] + 2 * s[i-1].

https://github.com/starcroce/leetcode/blob/master/roman_to_int.cpp

// 304 ms for 3999 test cases
class Solution {
public:
    int romanToInt(string s) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int ans = 0;
        for(int i = 0; i < s.size(); i++) {
            // like IV
            if(i > 0 && charToInt(s[i]) > charToInt(s[i-1])){
                ans += charToInt(s[i]) - 2 * charToInt(s[i-1]);
            }
            else {
                ans += charToInt(s[i]);
            }
        }
        return ans;
    }
    int charToInt(char c) {
        switch(c) {
            case 'I':
                return 1;
            case 'V':
                return 5;
            case 'X':
                return 10;
            case 'L':
                return 50;
            case 'C':
                return 100;
            case 'D':
                return 500;
            case 'M':
                return 1000;
            default:
                return 0;
        }
    }
};