Day018 - 12. Integer to Roman

12. Integer to Roman

Seven different symbols represent Roman numerals with the following values:

Symbol	Value
I	1
V	5
X	10
L	50
C	100
D	500
M	1000

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:

Input: num = 3749

Output: “MMMDCCXLIX”

Explanation:

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:

Input: num = 58

Output: “LVIII”

Explanation:

50 = L
 8 = VIII

Example 3:

Input: num = 1994

Output: “MCMXCIV”

Explanation:

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

Constraints:

• 1 <= num <= 3999

내 풀이

class Solution:
    def intToRoman(self, num: int) -> str:
        dict_1 = {1:"I", 4:"IV", 5:"V", 9:"IX"}
        dict_10 = {1:"X", 4:"XL", 5:"L", 9:"XC"}
        dict_100 = {1:"C", 4:"CD", 5:"D", 9:"CM"}

        num = list(map(int, list(str(num))))

        zero_mul = len(num)-1

        answer = ""

        for i in num:
            if zero_mul == 3: # 1000의자리
                answer += "M"*i
                zero_mul -= 1
                continue

            if i == 0:
                zero_mul -= 1
                continue

            else: 
                if zero_mul == 2:
                    use_dict = dict_100
                elif zero_mul == 1:
                    use_dict = dict_10
                else:
                    use_dict = dict_1

            if i in use_dict:
                answer += use_dict[i]
            else:
                if i < 4:
                    answer += use_dict[1] * i
                else:
                    answer += use_dict[5] + use_dict[1] * (i-5)

            zero_mul -= 1
                
        return answer

# Time Complexity : \(O(N)\)