#####solution1####faster####def romanToInt(s):    d={        'I':1,        'V':5,        'X':10,        'L':50,        'C':100,        'D':500,        'M':1000    }    i = 1    count = last = d[s[0]]    while i < len(s):        current = d[s[i]]        if current > last:            count -= last * 2        count += current        last = current        i += 1    return count# ########solution2######### def romanToInt(s):#     d = {'I': 1,#          'V': 5,#          'X': 10,#          'L': 50,#          'C': 100,#          'D': 500,#          'M': 1000#          }#     res = 0#     if len(s) < 2:#         return d[s[0]]#     res = res + d[s[0]]#     for i in range(1,len(s)):#         res = res + d[s[i]]#         if d[s[i]] > d[s[i - 1]]:#             res = res - 2 * d[s[i-1]]#     return res### if __name__=='__main__':#     m="MCMXCIV"#     print(romanToInt(m))

　　

　　分析：

current记录当前元素值，last记录前一个元素值，count记录current之前所有元素的和，也就是加上了last。

当current大于last时，count需要先减去last再加上current-last，即count-2*last+current

当current小于last时，count直接加上当前current即可，即count+current