#####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