题目如下：
题目
这道题提供两种思路，一种是用判断语句穷举所有情况，参考代码如下：
class Solution:
def romanToInt(self, s):
"""
:type s: str
:rtype: int
"""
roman = s
num = 0
rlen = len(roman)
if rlen == 0:
return 0
i = 0
while i < rlen:
if roman[i] == "M":
num = num + 1000
i = i + 1
continue
if roman[i] == "C":
if i + 1 < rlen:
if roman[i+1] == "M":
num = num + 900
i = i + 2
continue
elif roman[i+1] == "D":
num = num + 400
i = i + 2
continue
else:
num = num + 100
i = i + 1
continue
else:
num = num + 100
i = i + 1
continue
if roman[i] == "D":
num = num + 500
i = i + 1
continue
if roman[i] == "X":
if i + 1 < rlen:
if roman[i+1] == "C":
num = num + 90
i = i + 2
continue
elif roman[i+1] == "L":
num = num + 40
i = i + 2
continue
else:
num = num + 10
i = i + 1
continue
else:
num = num + 10
i = i + 1
continue
if roman[i] == "L":
num = num + 50
i = i + 1
continue
if roman[i] == "I":
if i + 1 < rlen:
if roman[i+1] == "X":
num = num + 9
i = i + 2
continue
elif roman[i+1] == "V":
num = num + 4
i = i + 2
continue
else:
num = num + 1
i = i + 1
continue
else:
num = num + 1
i = i + 1
continue
if roman[i] == "V":
num = num + 5
i = i + 1
continue
return num
上面这种解法代码有点繁复，下面提供另一种解法，通过对于的罗马数字字典来倒序处理字符串，值得注意的是，如果当前字符比上次迭代的字符所代表的数字要小，那么应该从总数中减去。如“CD，IV”。参考代码如下：
class Solution:
def romanToInt(self, s):
"""
:type s: str
:rtype: int
"""
roman_id = {
'I': 1,
'V': 5,
'X': 10,
'L': 50,
'C': 100,
'D': 500,
'M': 1000,
}
c = 0
num = 0
for i in range(len(s) - 1, -1, -1):
j = roman_id[s[i]]
# 如果当前字符小于上次迭代的字符，那么属于特殊请求（e.g.,CD,IV..）
if j < c:
num -= j
# 否则直接相加
if j >= c:
num += j
c = j
return num
如果您有更好的实现方法，欢迎交流。
ps：如果您有好的建议，欢迎交流 :-D，也欢迎访问我的个人博客：tundrazone.com
