[LeetCode] 12. Integer to Roman (Python)

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12. Integer to Roman

Medium

Seven different symbols represent Roman numerals with the following values:

Roman numerals are formed by appending the conversions of decimal place values from highest to lowest. Converting a decimal place value into a Roman numeral has the following rules:

• If the value does not start with 4 or 9, select the symbol of the maximal value that can be subtracted from the input, append that symbol to the result, subtract its value, and convert the remainder to a Roman numeral.
• If the value starts with 4 or 9 use the subtractive form representing one symbol subtracted from the following symbol, for example, 4 is 1 (I) less than 5 (V): IV and 9 is 1 (I) less than 10 (X): IX. Only the following subtractive forms are used: 4 (IV), 9 (IX), 40 (XL), 90 (XC), 400 (CD) and 900 (CM).
• Only powers of 10 (I, X, C, M) can be appended consecutively at most 3 times to represent multiples of 10. You cannot append 5 (V), 50 (L), or 500 (D) multiple times. If you need to append a symbol 4 times use the subtractive form.

Given an integer, convert it to a Roman numeral.

Example 1:

3000 = MMM as 1000 (M) + 1000 (M) + 1000 (M)
 700 = DCC as 500 (D) + 100 (C) + 100 (C)
  40 = XL as 10 (X) less of 50 (L)
   9 = IX as 1 (I) less of 10 (X)
Note: 49 is not 1 (I) less of 50 (L) because the conversion is based on decimal places

Example 2:

50 = L
 8 = VIII

Example 3:

1000 = M
 900 = CM
  90 = XC
   4 = IV

Constraints:

• 1 <= num <= 3999

Code

class Solution(object):
    def intToRoman(self, num):
        global roman_num
        roman = {1: 'I', 5: 'V', 10: 'X', 50: 'L', 100: 'C', 500: 'D', 1000: 'M'}
        n_list = list(reversed(list(map(int, str(num)))))
        roman_num = ""
        digits = sorted(list(roman.keys()))

        def func(n):
            global roman_num
            if n_list[n] == 1:
                roman_num += roman[digits[n*2]]
            elif n_list[n] == 2:
                roman_num += (roman[digits[n*2]] * 2)
            elif n_list[n] == 3:
                roman_num += (roman[digits[(n*2)]] * 3)
            elif n_list[n] == 4:
                roman_num += (roman[digits[(n*2)+1]] + roman[digits[n*2]])   
            elif n_list[n] == 5:
                roman_num += roman[digits[(n*2)+1]]
            elif n_list[n] == 6:
                roman_num += (roman[digits[n*2]] + roman[digits[(n*2)+1]])   
            elif n_list[n] == 7:
                roman_num += (roman[digits[n*2]] * 2 + roman[digits[(n*2)+1]])  
            elif n_list[n] == 8:
                roman_num += (roman[digits[n*2]] * 3 + roman[digits[(n*2)+1]])
            elif n_list[n] == 9:
                roman_num += (roman[digits[(n*2)+2]] + roman[digits[n*2]])
        
        for n in range(len(n_list)):
            if n == 3:
                roman_num += ('M' * n_list[n])
            else:
                func(n)
            
        return ''.join(reversed(roman_num))

무식하게 푼 방법이다.

예를 들어, 3749를 생각해보면, XILXCCDMMM이 나와야 한다.

내가 푼 방법은 n_list = [9, 4, 7, 3]로 역순으로 바꾼 리스트를 일일이 확인해주면서 dictionary에서 값을 찾아 적용해주었다.

Time Complexity