LintCode 428. The n-th power Python algorithm of x

description

Implement pow(x, n). (n is an integer)

Description

Don't worry about accuracy, it is correct when the absolute value of the difference between the answer and the standard output is less than 1e-3

Sample

- 样例 1:

输入: x = 9.88023, n = 3
输出: 964.498
- 样例 2:

输入: x = 2.1, n = 3
输出: 9.261

- 样例 3:

输入: x = 1, n = 0
输出: 1

challenge

Time complexity O(logn)

Parsing

class Solution:
    def myPow(self, x, n):
        if x == 0:
            return 0
            
        if n < 0:
            x = 1.0 / x
            n = abs(n)
        
        res = 1
        while n != 0:
            if n & 1:
                res *= x
            x *= x
            n = n >> 1
        return res

operation result