leetcode Pow(x, n) -- 简单重点

https://leetcode.com/problems/powx-n/

mycode：

class Solution(object):
    def myPow(self, x, n):
        """ :type x: float :type n: int :rtype: float """
        if n == 0:
            return 1.0
        elif n == 1:
            return x
        elif n < 0:
            return 1.0 / self.myPow(x, -n)
        elif n > 0 :
            if n % 2 == 0:
                return self.myPow( x, n/2) ** 2
            else:
                return self.myPow(x, n/2) ** 2 * x