leetcode/problems/0050-py/main.py at main · emfomy/leetcode · GitHub

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# Source: https://leetcode.com/problems/powx-n
# Title: Pow(x, n)
# Difficulty: Medium
# Author: Mu Yang <http://muyang.pro>

################################################################################################################################
# Implement **pow(x, n)**, which calculates `x` raised to the power `n` (i.e., `x^n`).
#
# Reference: https://cplusplus.com/reference/valarray/pow/
#
# **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 = 1/2^2 = 1/4 = 0.25
# ```
#
# **Constraints:**
#
# - `-100.0 < x < 100.0`
# - `-2^31 <= n <= 2^31-1`
# - `n` is an integer.
# - Either `x` is not zero or `n > 0`.
# - `-10^4 <= x^n <= 10^4`
#
################################################################################################################################


class Solution:
    def myPow(self, x: float, n: int) -> float:
        if x == 0.0:
            if n > 0:
                return 0.0
            if n == 0:
                return float("nan")
            if n < 0:
                return float("inf")

        if n < 0:
            n = -n
            x = 1.0 / x

        ans = 1.0
        while n > 0:
            if n % 2 == 1:
                ans *= x
            n //= 2
            x *= x

        return ans