[PYTHON] leet code easy Roman to Integer Approach 1: Passage de gauche à droite

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, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven 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. Input is guaranteed to be within the range from 1 to 3999.

class Solution:
    def romanToInt(self, s: str) -> int:
#Créez un dictionnaire de nombres et de nombres romains et recherchez dans une boucle while
        v ={"I":1, "V":5, "X":10, "L":50, "C":100, "D":500, "M":1000}
        total = 0
        i = 0
        while i < len(s):
#La partie difficile de ce problème est 4,Deux sur neuf. À d'autres moments, les nombres diminuent de gauche à droite. Par conséquent, dans ce qui suit, le branchement conditionnel est effectué avec ne serait-ce que pour les «exceptions où le nombre devient plus grand» de gauche à droite.
            if i + 1 < len(s) and v[s[i]] < v[s[i + 1]]:
#Au moment de l'exception(IX etc.)Est du nombre (X) à droite au nombre à gauche(I)Additionnez la valeur moins. Au total.
                total += v[s[i+1]] - v[s[i]]
                i += 2
            else:
                total += v[s[i]]
                i += 1
        return total