Leecode #13 Roman to Integer

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博客围绕LeetCode第十三题展开，该题要求将罗马数字转换为整数。介绍了罗马数字的表示规则，包括基本符号对应的值以及减法使用的六种情况，并给出多个示例。后续还将阐述解题思路和实现代码。

一、问题描述
Leecode第十三题，题目为：
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.

问题理解为：

二、解题思路

三、实现代码

class Solution {
public:
    int romanToInt(string s) {
        unordered_map<char, int> romanMap = { { 'I', 1 },
                                               { 'V', 5 },
                                               { 'X', 10 },
                                               { 'L', 50 },
                                               { 'C', 100 },
                                               { 'D', 500 },
                                               { 'M', 1000 } };
        int num = romanMap.at(s.back());
        for(int i=s.length()-2;i>=0;--i) {
            if(romanMap.at(s.at(i)) < romanMap.at(s.at(i+1)))
                num -= romanMap.at(s.at(i));
            else
                num += romanMap.at(s.at(i));
        }
        return num;
    }
};


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