LeetCode - Pow(x, n)

Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

 

Example 1:

Input: x = 2.00000, n = 10
Output: 1024.00000

Example 2:

Input: x = 2.10000, n = 3
Output: 9.26100

Example 3:

Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

 

Constraints:

• -100.0 < x < 100.0
• -231 <= n <= 231-1
• n is an integer.
• Either x is not zero or n > 0.
• -104 <= xn <= 104

 

My Solution:

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n == 0:
            return 1
        elif n == 1:
            return x
        elif n > 1:
            q, r = divmod(n, 2)
            y = self.myPow(x, q)
            return y * y * x ** r
        elif n < 0:
            return 1 / self.myPow(x, -1 * n)