自定义幂函数python计算x.的n次方_一个数number的n次幂 python的pow函数

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pow函数的实现——leetcode.

解法1：暴力法

不是常规意义上的暴力，过程中通过动态调整底数的大小来加快求解。代码如下：

def my_pow(number, n):

judge = True

if n < 0:

n = -n

judge = False

if n == 0:

return 1

result = 1

count = 1

temp = number

while n > 0:

if n >= count:

result *= temp

temp = temp * number

n -= count

count += 1

else:

temp /= number

count -= 1

return result if judge else 1/judge

解法2：根据奇偶幂分类(递归法，迭代法，位运算法)

如果n为偶数，则pow(x,n) = pow(x^2, n/2)；

如果n为奇数，则pow(x,n) = x*pow(x^2, (n-1)/2)。

class MyPow:

def my_pow(self, number, n):

if n < 0:

n = -n

return 1/self.help_(number, n)

return self.help_(number, n)

def help_(self, number, n):

if n == 0:

return 1

if n%2 == 0:

return self.help_(number*number, n//2)

return self.help_(number*number, (n-1)//2)*number

迭代代码如下：

class MyPow:

def my_pow(self, number, n):

judge = True

if n < 0:

n = -n

judge = False

result = 1

while n > 0:

if n%2 == 0:

number *= number

n //= 2

result *= number

n -= 1

return result if judge else 1/result

其实跟上面的方法类似，只是通过位运算符判断奇偶性并且进行除以2的操作(移位操作)。代码如下：

class Solution:

def myPow(self, x: float, n: int) -> float:

judge = True

if n < 0:

n = -n

judge = False

final = 1

while n>0:

if n & 1: #代表是奇数

final *= x

x *= x

n >>= 1 # 右移一位

return final if judge else 1/final