power()函数只适用于整数

int power(int base, unsigned int exp){

    if (exp == 0)
        return 1;
    int temp = power(base, exp/2);
    if (exp%2 == 0)
        return temp*temp;
    else
        return base*temp*temp;

}

复杂度= O(exp)

Power()函数为负exp和浮点基数工作。

float power(float base, int exp) {

    if( exp == 0)
       return 1;
    float temp = power(base, exp/2);       
    if (exp%2 == 0)
        return temp*temp;
    else {
        if(exp > 0)
            return base*temp*temp;
        else
            return (temp*temp)/base; //negative exponent computation 
    }

} 

复杂度= O(exp)

一种非常特殊的情况是，当你需要2^(-x ^ y)时，其中x当然是负的y太大了，不能对int型进行移位。你仍然可以用浮点数在常数时间内完成2^x。

struct IeeeFloat
{

    unsigned int base : 23;
    unsigned int exponent : 8;
    unsigned int signBit : 1;
};


union IeeeFloatUnion
{
    IeeeFloat brokenOut;
    float f;
};

inline float twoToThe(char exponent)
{
    // notice how the range checking is already done on the exponent var 
    static IeeeFloatUnion u;
    u.f = 2.0;
    // Change the exponent part of the float
    u.brokenOut.exponent += (exponent - 1);
    return (u.f);
}

使用double作为基底类型，可以得到更多的2的幂。 (非常感谢评论者帮助整理这篇文章)。

还有一种可能性是，学习更多关于IEEE浮点数的知识，其他幂运算的特殊情况可能会出现。

O(log N)的解决方案在Swift…

// Time complexity is O(log N)
func power(_ base: Int, _ exp: Int) -> Int { 

    // 1. If the exponent is 1 then return the number (e.g a^1 == a)
    //Time complexity O(1)
    if exp == 1 { 
        return base
    }

    // 2. Calculate the value of the number raised to half of the exponent. This will be used to calculate the final answer by squaring the result (e.g a^2n == (a^n)^2 == a^n * a^n). The idea is that we can do half the amount of work by obtaining a^n and multiplying the result by itself to get a^2n
    //Time complexity O(log N)
    let tempVal = power(base, exp/2) 

    // 3. If the exponent was odd then decompose the result in such a way that it allows you to divide the exponent in two (e.g. a^(2n+1) == a^1 * a^2n == a^1 * a^n * a^n). If the eponent is even then the result must be the base raised to half the exponent squared (e.g. a^2n == a^n * a^n = (a^n)^2).
    //Time complexity O(1)
    return (exp % 2 == 1 ? base : 1) * tempVal * tempVal 

}

平方求幂。

int ipow(int base, int exp)
{
    int result = 1;
    for (;;)
    {
        if (exp & 1)
            result *= base;
        exp >>= 1;
        if (!exp)
            break;
        base *= base;
    }

    return result;
}

这是在非对称密码学中对大数进行模求幂的标准方法。

我的情况有点不同，我试图用一种力量创造一个面具，但我想无论如何我都要分享我找到的解决方案。

显然，它只适用于2的幂。

Mask1 = 1 << (Exponent - 1);
Mask2 = Mask1 - 1;
return Mask1 + Mask2;

int pow( int base, int exponent)

{   // Does not work for negative exponents. (But that would be leaving the range of int) 
    if (exponent == 0) return 1;  // base case;
    int temp = pow(base, exponent/2);
    if (exponent % 2 == 0)
        return temp * temp; 
    else
        return (base * temp * temp);
}