[Leetcode] 13. Roman to Integer

題目 : 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

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.

1.

class Solution:
    def romanToInt(self, s: str) -> int:
        # 先建立羅馬數字的字典roman
        roman = {
            "I": 1,
            "V": 5,
            "X": 10,
            "L": 50,
            "C": 100,
            "D": 500,
            "M": 1000
        }
			   ​#字串是從最後面開始看，所以建立一個 s 的倒序字串 target，方便計算數字的總和(total)
        target = s[::-1]
        total = 0 
        if len(target) == 1:  # 如果長度是1就可以直接回傳該羅馬字母所代表的數字 
            total = roman[s]
            return total
        i = 0
        index = len(target)
        while(i<len(s)):  
        			#如果目前的羅馬數字>下一個羅馬字母的值，
        			#就可以將下一個羅馬字母的值減去目前的羅馬數字，再將i+2            
        			if i+1 < index and roman[target[i]] > roman[target[i+1]]:                
            		total = total + roman[target[i]] - roman[target[i+1]]
                i = i+2
            else:
            # 若是其他的狀況則是將目前羅馬字母的值加到total
                total = total + roman[target[i]]
                i = i+1
        return total