[LintCode] Roman to Integer 罗马数字转化成整数

Given a roman numeral, convert it to an integer.

The answer is guaranteed to be within the range from 1 to 3999.

Have you met this question in a real interview?

Yes

Clarification

What is Roman Numeral?

• https://en.wikipedia.org/wiki/Roman_numerals
• https://zh.wikipedia.org/wiki/%E7%BD%97%E9%A9%AC%E6%95%B0%E5%AD%97
• http://baike.baidu.com/view/42061.htm

Example

IV -> 4

XII -> 12

XXI -> 21

XCIX -> 99

LeetCode的原题，请参见我之前的博客Roman to Integer。

解法一：

class Solution {
public:
    /**
     * @param s Roman representation
     * @return an integer
     */
    int romanToInt(string& s) {
        int res = ;
        unordered_map<char, int> m{{'I', }, {'V', }, {'X', }, {'L', }, {'C', }, {'D', }, {'M', }};
        for (int i = ; i < s.size(); ++i) {
            if (i == s.size() -  || m[s[i]] >= m[s[i + ]]) res += m[s[i]];
            else res -= m[s[i]];
        }
        return res;
    }
};

解法二：

class Solution {
public:
    /**
     * @param s Roman representation
     * @return an integer
     */
    int romanToInt(string& s) {
        int res = ;
        unordered_map<char, int> m{{'I', }, {'V', }, {'X', }, {'L', }, {'C', }, {'D', }, {'M', }};
        for (int i = ; i < s.size(); ++i) {
            if (i == ) res += m[s[i]];
            else {
                if (m[s[i]] > m[s[i - ]]) res += m[s[i]] -  * m[s[i - ]];
                else res += m[s[i]];
            }
        }
        return res;
    }
};