Roman to integer

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文章讲述了如何将罗马数字转换为整数，主要方法包括使用Python的字典配合特殊规则处理减法规则，以及C++中的switch-case结构进行逐个字符判断。两种方法都考虑了罗马数字中如IV、IX等表示减法的特殊情况。

一、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, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 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.

二、代码

Answer 1. 可以使用python的字典来存储7个罗马字符，然后遍历字符串。

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        roman_map = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        roman_map_special = {'IV':4, 'IX':9, 'XL':40, 'XC':90, 'CD':400, 'CM':900}
        res = 0
        i = 0
        while i < Len(s):
            if i+1<len(s) and s[i:i+2] in roman_map_special:
                res += roman_map_special[s[i:i+2]]
                i +=2
            else:
                res += roman_map[s[i:i+2]]
                i +=1
        return res

Answer 2. C++：用switch-case or if- else if分成7种情况。

class Solution {
public:
    int romanToInt(string s) {
        int sum = 0;
        for (int i = 0; i < s.length(); i++){
            switch (s[i]){
                case 'I':
                    if (i < s.length() - 1 && (s[i+1]=='V'||s[i+1]=='X')){
                        sum -=1;
                    }else{sum +=1;}break;
                case 'V':
                    sum += 5;
                    break;
                case 'X':
                    if (i < s.length() - 1 && (s[i+1]=='L'||s[i+1]=='C')){
                        sum -=10;
                    }else{sum +=10;}break;
                case 'L':
                    sum +=50;break;
                case 'C':
                    if (i < s.length() - 1 && (s[i+1]=='D'||s[i+1]=='M')){
                        sum -=100;
                    }else{sum +=100;}break;
                case 'D':
                    sum +=500;
                    break;
                case 'M':
                    sum +=1000;
                    break;
            }
        }
        return sum;
    }
};