13. Roman to Integer

问题描述：

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 four 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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

解题思路：

不难看出：

　　1. 当当前数字比其右边数字小的时候，则减去当前数字。

　　2. 当当前数字比其右边数字大或等于右边数字时，则加上当前数字。

与12. Integer to Roman可构成一个系列

代码：

class Solution {
public:
    int romanToInt(string s) {
        if(s.size() == 0)
            return 0;
        unordered_map<char, int> charMap;
        charMap['I'] = 1;
        charMap['V'] = 5;
        charMap['X'] = 10;
        charMap['L'] = 50;
        charMap['C'] = 100;
        charMap['D'] = 500;
        charMap['M'] = 1000;
        int n = s.size()-1;
        int ret = charMap[s[n]];
        for(int i = n-1; i > -1; i--){
            if(charMap[s[i+1]] > charMap[s[i]])
                ret -= charMap[s[i]];
            else
                ret += charMap[s[i]];
        }
        return ret;
    }
};