50. Pow(x, n)
Problem
Implement pow(x, n), which calculates x raised to the power n (i.e., xⁿ).
https://leetcode.com/problems/powx-n/
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 <= 2³¹-1nis an integer.- Either
xis not zero orn > 0. -10⁴ <= xⁿ <= 10⁴
Test Cases
class Solution:
def myPow(self, x: float, n: int) -> float:solution_test.py下载完整源码
import pytest
from solution import Solution
@pytest.mark.parametrize('x, n, expected', [
(2.00000, 10, 1024.00000),
(2.10000, 3, 9.26100),
(2.00000, -2, 0.25000),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, x, n, expected):
assert sol.myPow(x, n) == pytest.approx(expected, abs=1e-6)
Thoughts
在 3266. Final Array State After K Multiplication Operations II 中已经实现了两种二分法幂运算的逻辑，直接套用其中一种即可。
如果 n 是负数，则令 x = 1 / x，然后计算 x⁻ⁿ 即可。
Code
solution.py下载完整源码
class Solution:
def myPow(self, x: float, n: int) -> float:
if n < 0: x, n = 1 / x, -n
res = 1
while n > 0:
if n & 1:
res *= x
n >>= 1
x *= x
return res
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