python--剑指offer--16. 数值的整数次方

实现函数double Power(double base, int exponent)，求base的exponent次方。不得使用库函数，同时不需要考虑大数问题。

示例 1:

输入: 2.00000, 10
输出: 1024.00000

示例 2:

输入: 2.10000, 3
输出: 9.26100

示例 3:

输入: 2.00000, -2
输出: 0.25000
解释: 2-2 = 1/22 = 1/4 = 0.25

示例 1:

输入: 2.00000, 10
输出: 1024.00000
示例 2:

输入: 2.10000, 3
输出: 9.26100
示例 3:

输入: 2.00000, -2
输出: 0.25000
解释: 2-2 = 1/22 = 1/4 = 0.25

说明:

-100.0 < x < 100.0
n 是 32 位有符号整数，其数值范围是 [−231, 231 − 1] 。

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if x == 0:
            return 0

        if n < 0:
            n = -n
            x = 1/x

        res = 1
        while n != 0:
            if n & 1:
                res *= x
            x *= x
            n >>= 1
        return round(res, 5)


if __name__ == '__main__':
    solution = Solution()
    a = 2.0000
    b = 10
    reslut = solution.myPow(a, b)
    print(reslut)

    a = 2.1000
    b = 3
    reslut = solution.myPow(a, b)
    print(reslut)

    a = 2.0000
    b = -2
    reslut = solution.myPow(a, b)
    print(reslut)


"""
运行结果：
1024.0
9.261
0.25

Process finished with exit code 0
"""

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