leetcode 13 Roman to Integer

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Roman to Integer

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

从字符串中每次处理一位，处理前需要先找到剩余串的枢纽（剩余串中最大的），以判断当前值在其左边还是右边，该相加还是相减。

import unittestfrom pprint import pprintimport pdbclass Solution:    # @return an integer    def romanToInt(self, s):        dic={}        dic["I"]=1        dic["V"]=5        dic["X"]=10        dic["L"]=50        dic["C"]=100        dic["D"]=500        dic["M"]=1000        lst=("I","V","X","L","C","D","M")        num=0        for i in range(len(s)):            for j in lst:                if j in s[i:]:                    pivot=dic[j] #determin add or sub            if dic[s[i]]>=pivot:                num+=dic[s[i]]            else:                num-=dic[s[i]]        return numclass testCase(unittest.TestCase):    def setUp(self):        pass        self.a=Solution()    def testLeet(self):        self.assertEqual(self.a.romanToInt("I"),1)        self.assertEqual(self.a.romanToInt("V"),5)        self.assertEqual(self.a.romanToInt("X"),10)        self.assertEqual(self.a.romanToInt("XXXIX"),39)        self.assertEqual(self.a.romanToInt("CD"),400)        self.assertEqual(self.a.romanToInt("MCM"),1900)        self.assertEqual(self.a.romanToInt("MDCLXVI"),1666)        self.assertEqual(self.a.romanToInt("MDCCCLXXXVIII"),1888)        self.assertEqual(self.a.romanToInt("MCMLXXXIV"),1984)if __name__ == '__main__':    unittest.main()

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