python实现leetcode-第13题-罗马数字转整数
class Solution:
def romanToInt(self, s):
"""
:type s: str
:rtype: int
"""
num1=num4=num5=num9=num10=num40=num50=num90=num100=num400=num500=num900=num1000=0
for i in range(2):
tmp = 0
Length=len(s)
for tmp in range(Length):
if s[tmp:tmp+2]=='IV':
num4=num4+1
s = s[:tmp]+s[tmp+2:]
elif s[tmp:tmp+2]=='IX':
num9=num9+1
s = s[:tmp]+s[tmp+2:]
elif s[tmp:tmp+2]=='XL':
num40=num40+1
s = s[:tmp]+s[tmp+2:]
elif s[tmp:tmp+2]=='XC':
num90=num90+1
s = s[:tmp]+s[tmp+2:]
elif s[tmp:tmp+2]=='CD':
num400=num400+1
s = s[:tmp]+s[tmp+2:]
elif s[tmp:tmp+2]=='CM':
num900=num900+1
s = s[:tmp]+s[tmp+2:]
else:
s = s
print(s)
for i,num in enumerate(s):
#print (i,num)
if num == 'I':
num1 = num1+1
if num == 'V':
num5 = num5+1
if num == 'X':
num10 = num10+1
if num == 'L':
num50 = num50+1
if num == 'C':
num100 = num100+1
if num == 'D':
num500 = num500+1
if num == 'M':
num1000 = num1000+1
#print(num4,num9,num40,num90,num400,num900)
num = 1*num1+5*num5+10*num10+50*num50+100*num100+500*num500+1000*num1000+4*num4+9*num9+40*num40+90*num90+400*num400+900*num900
return num 注意：
1 采用先判断特殊条件，之后再常规条件相加
2 enumerate函数用法
3 字符串切割与拼接
4 循环两次是因为两次才能覆盖所有情况，避免遗漏（通过测试了，具体原理需要研究）
附大佬代码：
class Solution:
def romanToInt(self, s):
"""
:type s: str
:rtype: int
"""
a = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
ans=0
for i in range(len(s)):
if i<len(s)-1 and a[s[i]]<a[s[i+1]]:
ans-=a[s[i]]
else:
ans+=a[s[i]]
return ans