Pow(x, n)

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

Solution

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

Complexity

• time: O(log n)
• space: O(log n)