50. Pow(x, n) (Medium) - TengnanYao/daily_leetcode GitHub Wiki

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