class Solution:
def romanToInt(self, s: str) -> int:
num_convert = {'I': 1, 'V': 5, 'X':10,'L':50,'C':100,'D':500,'M':1000}
s_len=len(s)
count=0
for i in range(s_len):
if s[i] in num_convert:
if i==s_len-1:
count+=num_convert[s[i]]
continue
elif s[i]=='I' and s[i+1]=='V':
count-=num_convert[s[i]]
continue
elif s[i]=='I' and s[i+1]=='X':
count-=num_convert[s[i]]
continue
elif s[i]=='X' and s[i+1]=='L':
count-=num_convert[s[i]]
continue
elif s[i]=='X' and s[i+1]=='C':
count-=num_convert[s[i]]
continue
elif s[i]=='C' and s[i+1]=='D':
count-=num_convert[s[i]]
continue
elif s[i]=='C' and s[i+1]=='M':
count-=num_convert[s[i]]
continue
else:
count+=num_convert[s[i]]
return count
s = "MCMXCIV"
sol = Solution().romanToInt(s)
print(sol)
标签：count,convert,continue,leedcode,整数,罗马数字,num,len,elif From： https://www.cnblogs.com/yyyjw/p/17936560.html