50. Pow(x, n) @Python @Go

原题

https://leetcode.cn/problems/powx-n/description/

思路

递归

复杂度

时间：O(log(n))
空间：O(1)

Python代码

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

Go代码

func myPow(x float64, n int) float64 {
	if n == 0 {
		return 1
	} else if n == 1 {
		return x
	} else if n < 0 {
		return 1 / myPow(x, -n)
	} else if n%2 != 0 {
		return myPow(x, n-1) * x
	} else {
		root := myPow(x, n/2)
		return root * root
	}
}