算法题/数值的整数次方

python3

#coding:utf-8
def Power(base,n):
result = 1.0
if n == 0:
return 1
elif n == 1:
return base
elif n < 0:
n1 = -n
for i in range(n1):
result *= base
return 1.0/result
else:
for i in range(n):
result *= base
return result
print(Power(2,-4))

面向对象

#coding:utf-8
class Solution:
def Power(self,base,exponent):
self.result = 1.0
if exponent == 0:
return 1
elif exponent == 1:
return base
elif exponent < 0:
exponent = -exponent
for i in range(exponent):
self.result *= base
return 1.0/self.result
else:
for i in range(exponent):
self.result *= base
return self.result

if __name__ == '__main__':
s = Solution()
print(s.Power(2,-4))