50. Pow(x, n)

• categories: Code, Interview Question, leetcode, Medium
• source: https://leetcode.com/problems/powx-n
• topics: Mathematics, Recursion

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

Idea

Use the recursion to reduce the number of multiplications

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