[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 = 0;
        unordered_map<char, int> m{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
        for (int i = 0; i < s.size(); ++i) {
            if (i == s.size() - 1 || m[s[i]] >= m[s[i + 1]]) 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 = 0;
        unordered_map<char, int> m{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
        for (int i = 0; i < s.size(); ++i) {
            if (i == 0) res += m[s[i]];
            else {
                if (m[s[i]] > m[s[i - 1]]) res += m[s[i]] - 2 * m[s[i - 1]];
                else res += m[s[i]];
            }
        }
        return res;
    }
};