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 and9.
• 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.

Example 5:

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

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题目大意：

  把罗马数字转化为阿拉伯数字。罗马数字能够用I, V, X, L, C, D 和M这七个数字来表示。注意：数字介于1到3999之间。

解题思路：

  emmmmmmmmmmmmmmmmmmm~~~~~~~~~~~~~~~~，题目是昨天的对换版

  不用昨天的解法了，考虑用一个无序容器将罗马数字和对应的阿拉伯数字成对存起来先，然后1 2 3 5 6 7 8这种都容易判断，所以主要就在于判断4 和 9，所以要判断两个相邻位的前一个数是否比后一个数小，如果是就证明这两位数是用来构成4 或 9的，其他的就直接加所代表数字就行。

解题代码：

class Solution {
public:
    int romanToInt(string s) {

        map<char,int> conv = { {'M', 1000},{'D', 500},{'C', 100},{'L', 50},{'X', 10},{'V', 5},{'I', 1} };

        int res = 0;
        size_t it = 0, size = s.size() - 1;
        while (it <= size) {
            if (conv[s[it]] < conv[s[it + 1]])
            {
                res += (conv[s[it + 1]] - conv[s[it]]);
                ++it;
            }
            else
            {
                res += conv[s[it]];
            }
            ++it;
        }
        return res;
    }
};