50. Pow(x, n)

求x的n次幂。

思路：递归，注意负数

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