LeetCode 13. Roman to Integer

https://leetcode.com/problems/roman-to-integer/#/description

Given a roman numeral, convert it to an integer.

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

• 字符串简单题，要搞清楚转换规则。如果当前字母比前一个大，则相减，比如IV = 5 - 1；否则就相加，比如VI = 5 + 1，II = 1 + 1。
• 罗马数字_百度百科
  • http://baike.baidu.com/link?url=JBuRWsGjAmYIlIhaPN_ywmIJBMrTWT6iKb2-WhqyTA6RqitOQuqnvQ2PHVfelAf00iGWWtgTzUjB3W4YMR0XWLfadA6YVi_s2J1aUgb-n1eBewvqGmyRpdH3VsVVs4q3

#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
    inline int map(const char inCh)
    {
        switch (inCh)
        {
            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;
        }
        
    }
    
    int romanToInt(string s)
    {
        int result = 0;
        
        for (size_t i = 0; i < s.size(); i ++) {
            if ((i > 0) && (map(s[i]) > map(s[i - 1]))) {
                result += map(s[i]) - 2 * map(s[i - 1]);
            } else {
                result += map(s[i]);
            }
        }
        
        return result;
    }
};

int main ()
{
    Solution testSolution;
    string sTest[] = {"XXI", "XXIX"};

    for (int i = 0; i < 2; i ++)
        cout << testSolution.romanToInt(sTest[i]) << endl;
    
    return 0;
}