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