題目：

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value 
 I             1 
 V             5 
 X             10 
 L             50 
 C             100 
 D             500 
 M             1000 

For example,two is written as II in Roman numeral, just two one’s added together.Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II. Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number our is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used: I can be placed before V (5) and X (10) to make 4 and 9. X can be placed before L (50) and C (100) to make 40 and 90. C can be placed before D (500) and M (1000) to make 400 and 900. Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III" 
Output: 3 

Example 2:

Input: "IV" 
Output: 4 

Example 3:

Input: "IX" 
Output: 9

Example 4:

Input: "LVIII" 
Output: 58 
Explanation: L = 50, V= 5, III = 3. 

Example5:

Input: "MCMXCIV" 
Output: 1994 
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

解釋： 把羅馬數字轉換為十進位制整數。小數字出現在大數字之前時表示減法。 python程式碼：

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        roman_dic={'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
        sum=0
        for i in xrange(0,len(s)-1):
            if roman_dic[s[i]]<roman_dic[s[i+1]]:
                sum
 
-=roman_dic[s[i]]
            else:
                sum+=roman_dic[s[i]]
        sum+=roman_dic[s[-1]]
        return sum

c++程式碼：

#include<map>
#include<string>
using namespace std;
class Solution {
public:
    int romanToInt(string s) {
        map<char,int>_map;
        _map['I']=1;
        _map['V']=5;
        _map['X']=10;
        _map['L']=50;
        _map['C']=100;
        _map['D']=500;
        _map['M']=1000;
        int sum=0;
        for (int i=0;i<s.size()-1;i++)
        {
            if (_map[s[i]]<_map[s[i+1]])
                sum-=_map[s[i]];
            else
                sum+=_map[s[i]];
        }
        sum+=_map[s[s.size()-1]];
        return sum; 
    }
};

總結： c++好像確實沒有構造map的簡便方法。