[LeetCode][Python3] 50. Pow(x, n)

Problem :

https://leetcode.com/problems/powx-n/

My Solution :

class Solution:

def myPow(self, x: float, n: int) -> float:

if n < 0:

x = 1/x

n = -n

if n == 0:

return 1

if n == 1:

return x

if n % 2 == 0:

return self.myPow(x*x, n/2)

return x*self.myPow(x*x, (n-1)/2)

Comment :

이진 탐색처럼 곱하기 횟수를 O(logn)으로 줄여야 Time Limit Exceeded를 피할 수 있다.