[leetcode] - 13. Roman to Integer

13. Roman to Integer

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

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

For example, 2 is written as II in Roman numeral, just two ones 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. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer.

Example 1:

Input: s = "III"
Output: 3
Explanation: III = 3

Example 2:

Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 3:

Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

풀이

주어진 세가지 조건에 맞게 subtract를 할지 add를 할지 결정하여 풀었다.

class Solution(object):
    def romanToInt(self, s):

        symbol_map = {
            'I': 1,
            'V': 5,
            'X': 10,
            'L': 50,
            'C': 100,
            'D': 500,
            'M': 1000
        }

        answer = 0
        if len(s) == 1:
            return symbol_map[s]

        for i in range(len(s) - 1):
            if (s[i] == 'I' and s[i + 1] in ['V', 'X']) or (s[i] == 'X' and s[i + 1] in ['L', 'C']) or (s[i] == 'C' and s[i + 1]) in ['D', 'M']:
                answer -= symbol_map[s[i]]
            else:
                answer += symbol_map[s[i]]

        return answer + symbol_map[s[-1]]

답은 맞지만, 썩 좋은 풀이는 아닌 것 같아서 다른 사람들의 풀이를 참고 해야겠다.

다른 풀이 1:

조건을 간단하게 앞에 나온 숫자가 뒤에 나온 숫자보다 작다면, subtract를 하도록 설정했다.

class Solution:
    def romanToInt(self, s: str) -> int:
        m = {
            'I': 1,
            'V': 5,
            'X': 10,
            'L': 50,
            'C': 100,
            'D': 500,
            'M': 1000
        }

        ans = 0

        for i in range(len(s)):
            if i < len(s) - 1 and m[s[i]] < m[s[i+1]]:
                ans -= m[s[i]]
            else:
                ans += m[s[i]]

        return ans

다른 풀이 2:

나랑 동일하게 모든 경우의 수를 고려하여 풀었지만, .replace를 사용한 것이 참신했다.

class Solution:
    def romanToInt(self, s: str) -> int:
        translations = {
            "I": 1,
            "V": 5,
            "X": 10,
            "L": 50,
            "C": 100,
            "D": 500,
            "M": 1000
        }
        number = 0
        s = s.replace("IV", "IIII").replace("IX", "VIIII")
        s = s.replace("XL", "XXXX").replace("XC", "LXXXX")
        s = s.replace("CD", "CCCC").replace("CM", "DCCCC")
        for char in s:
            number += translations[char]
        return number