📌Problem

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 an integer, convert it to a roman numeral.

Example 1:

Input: num = 3
Output: "III"

Example 2:

Input: num = 4
Output: "IV"

Example 3:

Input: num = 9
Output: "IX"

Example 4:

Input: num = 58
Output: "LVIII"

Explanation: L = 50, V = 5, III = 3.

Example 5:

Input: num = 1994
Output: "MCMXCIV"
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].

leetcode 에서 풀기

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📝Solution

class Solution:
    def romanToInt(self, s: str) -> int:
        
        roman2value={'M':1000,'CM':900,'D':500,'CD':400,'C':100,'XC':90,'L':50,'XL':40,'X':10,'IX':9,'V':5,'IV':4,'I':1}
            
        value=0
        temp=''
        cursor=0
        while cursor<len(s):
            if (cursor+1)!=len(s) and s[cursor]+s[cursor+1] in roman2value:  # Check current character and next character
                value+=roman2value[s[cursor]+s[cursor+1]]
                cursor+=2
            else:
                value+=roman2value[s[cursor]]
                cursor+=1
                
            
        
        return value

• Time complexity : $O(N)$
• cursor를 이용하여 현재 문자와 다음 문자 cursor+1 을 체크해서 s[cursor]+s[cursor+1] 문자열이 roman2value 에 있다면 roman -> value 후 cursor를 두 칸 이동한다.
• 없다면 s[cursor]를 roman -> value 후 cursor를 한 칸 이동한다.