[Python-Leetcode] 13. Roman to Integer (difficulty : Easy - ☆)

문제 설명

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 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. 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.

 

Constraints:

• 1 <= s.length <= 15
• s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
• It is guaranteed that s is a valid roman numeral in the range [1, 3999].

 

 

문제 풀이

 

class Solution(object):
    def __init__(self):

        # 문자열 dict
        self.str_set = {'I': 1, 'V': 5, 'X': 10,
                        'L': 50, 'C': 100, 'D': 500, 'M': 1000}

    def romanToInt(self, s):
        self._sum = 0
        for i in range(len(s)-1):
            num = self.str_set[s[i]]
            next_num = self.str_set[s[i+1]]
        
        # 처음 문자의 숫자깂 >= 다음 문자의 숫자값
            if num >= next_num:
                self._sum += num
        
        # 처음 문자의 숫자깂 < 다음 문자의 숫자값
            else:
                self._sum -= num

        # len(s)-1 까지 확인하였으니, 맨 마지막 문자에 대한 숫자 확인 필요
        last_num = self.str_set[s[-1]]

        return self._sum + last_num