一个数number的n次幂 python的pow函数

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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

python位运算符简介.

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