leetcode50 Pow(x, n)

 1 """
 2 Implement pow(x, n), which calculates x raised to the power n (xn).
 3 Example 1:
 4 Input: 2.00000, 10
 5 Output: 1024.00000
 6 Example 2:
 7 Input: 2.10000, 3
 8 Output: 9.26100
 9 Example 3:
10 Input: 2.00000, -2
11 Output: 0.25000
12 Explanation: 2-2 = 1/22 = 1/4 = 0.25
13 """
14 class Solution:
15     def myPow(self, x, n):
16         if n < 0:
17             x = 1 / x  # 巧妙的转换了n<0的情况
18             n = -n
19         res = 1.0
20         while n:
21             if n % 2 != 0:  # n & 1  与运算高级写法，快了一点
22                 res *= x
23         x *= x
24         n = n // 2  # n >>= 1
25         return res