LeetCode-50 Pow(x, n)

Description

Implement pow(x, n), which calculates x raised to the power n (xⁿ).

Note: -100.0 < x < 100.0 n is a 32-bit signed integer, within the range [−2³¹, 2³¹ − 1]

Example

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

Submissions

这道题的解题思路主要采用递归计算,主要需要判断n的奇偶性和正负性.每次将n进行折半递归,当n为偶数时直接返回递归结果,当n为奇数时返回递归结果乘上x.且当n为正数时返回递归结果,x为负数时,递归结果取倒数.

实现代码如下:

class Solution: def myPow(self, x: float, n: int) -> float: if n == 0: return 1 if n == 1: return x if n == -1: return 1 / x if n%2 == 0: return self.myPow(x*x, n//2) else: return self.myPow(x*x, n//2) * x