428.x的n次幂

实现 pow(x,n)

不用担心精度，当答案和标准输出差绝对值小于1e-3时都算正确

样例

Pow(2.1, 3) = 9.261

Pow(0, 1) = 0

Pow(1, 0) = 1

挑战

O(logn) time

class Solution:
    """
    @param x: the base number
    @param n: the power number
    @return: the result
    """
    def myPow(self, x, n):
        if n==0:
            return 1

        if n<0:
            return 1/self.myPow(x,-n)

        elif n%2==0:
            return self.myPow(x*x,n//2)

        elif n%2==1:
            return x*self.myPow(x*x,n//2)