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罗马数字与阿拉伯数字转换

2019-04-11 10:49:29来源：博客园阅读 ()

罗马数字与阿拉伯数字对应关系如下：

且“II”表示2，“III”表示3，“IV”表示4，“VI表示6”，“VII”表示7，“VIII”表示8，“IX”表示9，其余的类似。

阿拉伯数转换成罗马数字

class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        
        if not num:
            return ""
        out = ""
        i = 3
        while i >= 0:
            out += self.get_roman(i,num//(10**i))
            num %= (10**i)
            i -= 1      
        return out
    
    def get_roman(self,power,quotient):
        power_to_roman = {0:["I","V","X"],1:["X","L","C"],2:["C","D","M"],3:["M"]}
        romans = power_to_roman[power]
        if quotient <= 3:
            out = quotient*romans[0]
        elif quotient == 4:
            out = romans[0]+romans[1]
        elif quotient == 5:
            out = romans[1]
        elif quotient <= 8:
            out = romans[1]+(quotient-5)*romans[0]
        else:
            out = romans[0]+romans[2]
        return out

罗马数字转换为阿拉伯数字：

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        if not s:
            return 0
        Roman_to_num = {'I':1,"V":5,"X":10,"L":50,"C":100,"D":500,"M":1000}
        before = {"V":"I","X":"I","L":"X","C":"X","D":"C","M":"C"}
        
        stack = []
        num = 0
        i = len(s)-1
        while i >= 0:
            if not stack:
                stack.append(s[i])
            else:
                last = stack.pop()
                if last in before and s[i] == before[last]:
                    num += Roman_to_num[last] - Roman_to_num[s[i]]
                else:
                    stack.append(last)
                    stack.append(s[i])
            i -= 1
        for i in stack:
            num += Roman_to_num[i]
        return num