13. Roman to Integer

Description

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.

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].

💡 Hint 1:
Problem is simpler to solve by working the string from back to front and using a map.

Submitted Code

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        hashmap = {}
        hashmap['I'] = 1    # 딕셔너리에 각 문자열을 키값으로 값 저장
        hashmap['V'] = 5
        hashmap['X'] = 10
        hashmap['L'] = 50
        hashmap['C'] = 100
        hashmap['D'] = 500
        hashmap['M'] = 1000

        reversed_c = ''     # 문자열 c를 거꾸로 세면서 보관하는 변수
        output = 0          # 출력값을 0으로 초기화

        for c in s[::-1]:
            if reversed_c == '':
                output += hashmap[c]
            elif c == 'I' and reversed_c[-1] == 'V':
                output -= 1
            elif c == 'I' and reversed_c[-1] == 'X':
                output -= 1
            elif c == 'X' and reversed_c[-1] == 'L':
                output -= 10
            elif c == 'X' and reversed_c[-1] == 'C':
                output -= 10
            elif c == 'C' and reversed_c[-1] == 'D':
                output -= 100
            elif c == 'C' and reversed_c[-1] == 'M':
                output -= 100
            else:
                output += hashmap[c]
            reversed_c += c
        return output

Runtime: 12 ms | Beats 40.02%
Memory: 12.52 MB | Beats 14.91%

문자열의 맨 뒤 문자부터 하나씩 변수 reversed_c에 보관하고,
I, X, C가 나올 때 reversed_c의 맨 마지막 원소를 확인하여 어떤 값을 output에 넣을지 결정했다.

Other Solutions

1st

class Solution:
    def romanToInt(self, s: str) -> int:
        res = 0
        roman = {
            'I': 1,
            'V': 5,
            'X': 10,
            'L': 50,
            'C': 100,
            'D': 500,
            'M': 1000
        }

        for a, b in zip(s, s[1:]):
            if roman[a] < roman[b]:
                res -= roman[a]
            else:
                res += roman[a]

        return res + roman[s[-1]]   # 마지막 남은 1개의 문자는 그대로 더하기

time complexity: 𝑂(𝑛) ← 입력 문자열 s의 길이
space complexity: 𝑂(1) ← s와 상관없이 딕셔너리의 크기는 동일

s = “XIV”
res = 0

문자를 앞에서부터 2개씩 묶어가며 계산하는 방식
두 문자 중 앞의 문자가 더 크면 앞의 문자를 합계에 더하고, 뒤의 문자가 더 크면 앞의 문자를 합계에서 뺀다.

loop	characters	result	res
1	X > I	+10	10
2	I < V	-1	9
3	V	+5

return res + roman[s[-1]]
∴ return 9 + 5

∴ return 14

2nd

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

        # Chaining 방식으로 문자열 변경 e.g. 먼저 s의 IV를 IIII로 변경 후, 그 값에 IX가 있을 경우 VIIII로 변경
        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

특수한 규칙을 변환 처리하여 계산을 단순화한 코드