[파이썬 Python/알고리즘] LeetCode - Integer to Roman

//GitHub https://github.com/vellimole0621

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12. Integer to Roman

https://leetcode.com/problems/integer-to-roman/

Integer to Roman - LeetCode

문제 설명

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Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

For example, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 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. 

Given an integer, convert it to a roman numeral.

=> 주어진 숫자를 로마자로 변환하라.

 

내 풀이

=> 노가다로 각각 큰 자리수 순으로 찾음

class Solution:
    def intToRoman(self, num: int) -> str:
        num_box = []

        if num >= 1000:
            num_thousand = int(num / 1000)
            for i in range(0, num_thousand):
                num_box.append("M")

        if num >= 100:
            if num >= 1000:
                num_hundred = int((num / 100) % 10)
            else:
                num_hundred = int(num / 100)

            if num_hundred == 9:
                num_box.append("CM")
            elif num_hundred == 4:
                num_box.append("CD")
            elif num_hundred > 5 and num_hundred != 9:
                num_box.append("D")
                for j in range(0, num_hundred-5):
                    num_box.append("C")
            elif num_hundred == 5:
                num_box.append("D")
            else:
                for k in range(0, num_hundred):
                    num_box.append("C")

        if num >= 10:
            if num >= 100:
                num_ten = int((num / 10) % 10)
            else:
                num_ten = int(num / 10)

            if num_ten == 9:
                num_box.append("XC")
            elif num_ten == 4:
                num_box.append("XL")
            elif num_ten > 5 and num_ten != 9:
                num_box.append("L")
                for j in range(0, num_ten - 5):
                    num_box.append("X")
            elif num_ten == 5:
                num_box.append("L")
            else:
                for k in range(0, num_ten):
                    num_box.append("X")

        if num >= 1:
            if num >= 10:
                num_one = int(num% 10)
            else:
                num_one = num

            if num_one == 9:
                num_box.append("IX")
            elif num_one == 4:
                num_box.append("IV")
            elif num_one > 5 and num_one != 9:
                num_box.append("V")
                for j in range(0, num_one - 5):
                    num_box.append("I")
            elif num_one == 5:
                num_box.append("V")
            else:
                for k in range(0, num_one):
                    num_box.append("I")

        ans = ''.join(num_box)
        return ans

 

 너무 성능이 떨어지는 풀이여서. 더 효과적인 풀이를 찾아보았다.

 

출처

[Python 으로 푸는 Leetcode]12. Integer to Roman

https://velog.io/@kgh732/Python-%EC%9C%BC%EB%A1%9C-%ED%91%B8%EB%8A%94-Leetcode12.-Integer-to-Roman

[Python 으로 푸는 Leetcode]12. Integer to Roman

class Solution:
    def intToRoman(self, num: int) -> str:
        
        value2roman={1000:'M',900:'CM',500:'D',400:'CD',100:'C',90:'XC',50:'L',40:'XL',10:'X',9:'IX',5:'V',4:'IV',1:'I'}
            
        roman=''
        while num!=0:
            for value in value2roman:
                if num-value >=0:
                    roman+=value2roman[value]
                    num=num-value
                    break
        
        return roman

 알고리즘 풀이 전, 접근 방식에 대해서 충분히 고려하고 풀이를 시작해야겠다...