0. 题目

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.

1. C++版本

思想就是先计算连续重复的字符，统计个数。在根据在V,X,L,C,D,M的左边还是右边，判断是+,还是-。运行耗时32 ms

    int romanToInt(string s) {
         std::map<char, int> rule{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
        int sum = 0, index = 0, duplicate = 1;
        char curr = s[index], next;
        for (;index < s.size()-1;) {
            next = s[index+1];
            while(curr == next) {
                duplicate++;
                index++;
                next = s[index+1];
            }
            if ((curr == 'I' && (next == 'V' || next == 'X') || (curr == 'X' && (next == 'L' || next == 'C')) ||(curr == 'C' && (next == 'D' || next == 'M')))) {
                sum -= rule[curr] * duplicate;
            }
            else {
                sum += rule[curr] * duplicate;
            }
            curr = next;
            duplicate = 1;
            index++;
        }
        sum += rule[curr] * duplicate;
        return sum;
    }

2. python版本

    def romanToInt(s):
        """
        :type s: str
        :rtype: int
        """
        rule = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        index, sum, duplicate = 0, 0, 1
        length = len(s)
        curr = s[index]
        while index < length-1:
             next = s[index+1]
             while index < length-2 and curr == next:
                 duplicate += 1
                 index += 1
                 next = s[index+1]
                
             if (curr == 'I' and (next =='V' or next == 'X')) or (curr == 'X' and (next == 'L' or next == 'C')) or (curr == 'C' and (next == 'D' or next == 'M')):
                 sum -= rule[curr] * duplicate
             else:
                 sum += rule[curr] * duplicate
             curr = next
             duplicate = 1
             index += 1
        if index == length-1:
            sum += rule[curr] * duplicate
        return sum