题目：
题目
这道题给出两种解法 ：
一种解法是通过条件判断求解，代码如下：
'''
Created on: Saturday, September 01, 2018
@author: Jedi Liu
'''
class Solution:
def intToRoman(self, num):
"""
:type num: int
:rtype: str
"""
n = num
roman = ''
m = n // 1000
if m > 0:
n = n - m * 1000
while m > 0:
m = m - 1
roman = roman + 'M'
c = n // 100
if c > 0:
if c == 9:
n = n - 900
roman = roman + 'CM'
elif c >= 5:
n = n - 500
roman = roman + 'D'
c = c - 5
n = n - c * 100
while c > 0:
c = c - 1
roman = roman + 'C'
elif c == 4:
n = n - 400
roman = roman + 'CD'
elif c > 0:
n = n - c * 100
while c > 0:
c = c - 1
roman = roman + 'C'
x = n // 10
if x > 0:
if x == 9:
n = n - 90
roman = roman + 'XC'
elif x >= 5:
n = n - 50
roman = roman + 'L'
x = x - 5
n = n - x * 10
while x > 0:
x = x - 1
roman = roman + 'X'
elif x == 4:
n = n - 40
roman = roman + 'XL'
elif x > 0:
n = n - x * 10
while x > 0:
x = x - 1
roman = roman + 'X'
if n == 9:
roman = roman + 'IX'
elif n >= 5:
n = n - 5
roman = roman + 'V'
while n > 0:
n = n - 1
roman = roman + 'I'
elif n == 4:
roman = roman + 'IV'
else:
while n > 0:
roman = roman + 'I'
n = n - 1
return roman
s = Solution()
print(s.intToRoman(27))
这种方法效率有点差，代码也不够简练，但容易想到。下面给出第二种解法，这种解法是利用罗马数字和阿拉伯数字的对应关系求解，代码如下：
'''
Created on: Saturday, September 01, 2018
@author: Jedi Liu
'''
class Solution1:
def intToRoman(self, num):
"""
:type num: int
:rtype: str
"""
n = num
roman = ""
div = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
romans = [
"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV",
"I"
]
while n != 0:
for i in range(len(div)):
d = div[i]
times = n // d
if times != 0:
roman = roman + (times) * romans[i]
n = n - d * times
return roman
如果你有更好的实现方法，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
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