leetcode50 Pow(x, n)

"""
Implement pow(x, n), which calculates x raised to the power n (xn).
Example 1:
Input: 2.00000, 10
Output: 1024.00000
Example 2:
Input: 2.10000, 3
Output: 9.26100
Example 3:
Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
"""
class Solution:
    def myPow(self, x, n):
        if n < 0:
            x = 1 / x  # 巧妙的转换了n<0的情况
            n = -n
        res = 1.0
        while n:
            if n % 2 != 0:  # n & 1  与运算高级写法，快了一点
                res *= x
        x *= x
        n = n // 2  # n >>= 1
        return res