题解 | #数值的整数次方#

# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param base double浮点型 
# @param exponent int整型 
# @return double浮点型
#
class Solution:
    def Power(self , base: float, exponent: int) -> float:
        res = 1
        for i in range(abs(exponent)):
            res *= base
        if exponent > 0:
            return res
        else:
            return 1/res

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param base double浮点型 
# @param exponent int整型 
# @return double浮点型
#
class Solution:
    def Power(self , base: float, exponent: int) -> float:
        if exponent >= 0:
            return self.PowerPositiveE(base, exponent)
        else:
            return 1.0 / self.PowerPositiveE(base, -exponent)
    def PowerPositiveE(self,base,exponent):
        if exponent == 0:
            return 1
        if exponent == 1:
            return base
        res = self.PowerPositiveE(base, exponent>>1)
        res *= res
        if (exponent & 0b1) == 1:
            res *= base
        return res