LeetCode第13题：Roman to Integer(C++)详解

Roman to Integer
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: L = 50, V= 5, III = 3.

Example 5:

Input: “MCMXCIV”
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
题目解析：罗马数值转换成阿拉伯数值需要找到其中的规律，规律如下：
1、基本数字Ⅰ、X 、C 中的任何一个，自身连用构成数目，或者放在大数的右边连用构成数目，都不能超过三个；放在大数的左边只能用一个。
2、不能把基本数字V 、L 、D 中的任何一个作为小数放在大数的左边采用相减的方法构成数目；放在大数的右边采用相加的方式构成数目，只能使用一个。
3、V 和X 左边的小数字只能用Ⅰ。
4、L 和C 左边的小数字只能用X。
5、D 和M 左边的小数字只能用C。
对于本题来说：
第一，如果当前数字是最后一个数字，或者之后的数字比它小的话，则加上当前数字。
第二，其他情况则减去这个数字。
此段规律总结来自于博客园大神：Grandyang，我就不再赘述，感谢大神总结。

题目代码：

class Solution {
public:
    int romanToInt(string s) {
        int result = 0;
        unordered_map romanMap{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
        for (int i = 0; i < s.size(); i++)
         {
            int val = romanMap[s[i]];
            if (i == s.size() - 1 || romanMap[s[i+1]] <= romanMap[s[i]]) 
            {
                 result += val;//查看上述规律
            }
            else 
            {
                 result -= val;
            }
        }
        return result;
    }
};

以LVII为例：
i=0时，romanMap[s[0]] = 50；此时romanMap[s[1]] = 5，则相加 50 + 5 = 55；
第二次循环i = 1；romanMap[s[1]] = 5，romanMap[s[2]] = 1，则相加 55 + 1 = 56;
第三次循环i = 2；romanMap[s[2]] = 1，romanMap[s[3]] = 1，则相加 56 + 1 = 57;
第三次循环i = 3；romanMap[s[3]] = 1，romanMap[s[4]] = 1，则相加 57 + 1 = 58.
性能：

总结：就是找到转换的相应规律，就可得到解题思路。