打卡群刷题总结0709—— Pow(x, n)

class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        if n == 0:
            return 1

        # 取得正负符号
        negetive = False
        if n < 0:
            negetive = True
            n = -n

        # 得到下标为2^i对应的数
        times = [0, 1]
        res = [1, x]
        while 2 * times[-1] < n:
            times.append(times[-1] * 2)
            res.append(res[-1] * res[-1])

        # 累积
        result = 1
        index = len(times) - 1
        while n > 0:
            while times[index] > n:
                index -= 1
            result *= res[index]
            n -= times[index]

        if negetive:
            return 1.0 / result
        return result