LeetCode 13 Roman to Integer

难度：easy

代码：

# -*- coding: utf-8 -*-
# @File  : leet_13.py
# @Author: FanyFull
# @Date  : 2021/9/25

class Solution:
    def romanToInt(self, s: str) -> int:
        roman_dict = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
        prenum = roman_dict[s[0]]
        num = 0
        for i in range(1, len(s)):
            single_num = roman_dict[s[i]]
            if prenum < single_num:
                num = num - prenum
            else:
                num = num + prenum
            prenum = single_num
        num = num + prenum
        return num

if __name__ == '__main__':
    solution = Solution()
    input = 'MCMXCIV'
    output = solution.romanToInt(input)
    print(output)

这道题的思想也可以说是加减交替吧，之所以说“也”，是因为最近上计算机组成原理也提到过这个说法。