13. 罗马数字转整数

symbolValues = { ‘I’: 1, ‘V’: 5, ‘X’: 10, ‘L’: 50, ‘C’: 100, ‘D’: 500, ‘M’: 1000 }
d = {‘I’: 1, ‘IV’: 3, ‘V’: 5, ‘IX’: 8, ‘X’: 10, ‘XL’: 30, ‘L’: 50, ‘XC’: 80, ‘C’: 100, ‘CD’: 300, ‘D’: 500, ‘CM’: 800, ‘M’: 1000}

class Solution:
    def romanToInt(self, s: str) -> int:
        # 统一正负，前面的字符加过了，双字符减去前面的。
        d = {'I':1, 'IV':3, 'V':5, 'IX':8, 'X':10, 'XL':30, 'L':50, 'XC':80, 'C':100, 'CD':300, 'D':500, 'CM':800, 'M':1000}
        return sum(d.get(s[max(i-1, 0):i+1], d[c]) for i, c in enumerate(s))
		# max[i-1,0] 处理首元素 i-1 = -1 的情况，d[c] 是默认值，d.get(s[max(i-1, 0):i+1], d[c])，处理双字符的情况。注意 双字符的值已经考虑前面字符加过的情况。
		# 考虑正负，单独加减。
        sv = { 'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000 }
        return sum(sv[c]*(-1) if i+1 < len(s) and sv[c] < sv[s[i+1]] else sv[c] for i, c in enumerate(s))

map = {‘I’: 1, ‘IV’: 4, ‘V’: 5, ‘IX’: 9, ‘X’: 10, ‘XL’: 40, ‘L’: 50, ‘XC’: 90, ‘C’: 100, ‘CD’: 400, ‘D’: 500, ‘CM’: 900, ‘M’: 1000}

遍历 s，如果双字符在 map 中，则将其值加到 ans，并后移 2 位。不在将当前字符值加到 ans，后移 1 位，最后返回 ans。

class Solution:
    def romanToInt(self, s: str) -> int:
        map = {'I': 1, 'IV': 4, 'V': 5, 'IX': 9, 'X': 10, 'XL': 40, 'L': 50, 'XC': 90, 'C': 100, 'CD': 400, 'D': 500, 'CM': 900, 'M': 1000}

        ans, n, i = 0, len(s), 0
        while i < n:   
            if i+1 < n and s[i:i+2] in map:
                ans += map[s[i:i+2]]
                i += 2
            else:
                ans += map[s[i]]
                i += 1
 
        return ans

SYMBOL = { ‘I’: 1, ‘V’: 5, ‘X’: 10, ‘L’: 50, ‘C’: 100, ‘D’: 500, ‘M’: 1000 }

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

        ans, n = 0, len(s)
        for i in range(1, n):
            v = SYMBOL[s[i-1]]
            ans += v*(-1) if v < SYMBOL[s[i]] else v
                
        return ans + SYMBOL[s[-1]]