数值的整数次方

class Solution {
public:
    double Power(double base, int exponent) {
        double res=1.0;
        if(base==0) return 0;
        if(exponent==0) return 1;
        if(exponent<0){
            base = 1.0/base;
            exponent = -exponent;
        } 
        for(int i=0;i<exponent;i++){
            res = res*base;
        }
        return res;
    }
};

递归
class Solution {
public:
    double Power(double base, int exponent) {
         
        if(exponent == 0) return 1;
        if(base == 0) return 0;
        if(exponent == 1)
            return base;
        else if(exponent == -1)
            return 1/base;
        return Power(base,exponent/2) * Power(base,exponent/2) * Power(base,exponent%2);
    }
};

class Solution {
public:
    double Power(double base, int exponent) {
        if(exponent>0)
        {
            if(exponent==1)
                return base;
            if(exponent%2==0)
                return Power(base,exponent/2)*Power(base,exponent/2);
            else
                return Power(base,exponent/2)*Power(base,exponent/2+1);
        }
        else if (exponent==0)
        {
            return 1;
        }
        else
        {
            return 1/Power(base,0-exponent);
        }
    }
};

快速幂

class Solution {
public:
    double Power(double base, int exponent) {
        long long p=abs((long long) exponent);
        double r = 1;
        if(base==0) return 0;
        while(p){
            if(p&1) r *= base;
            base *= base;
            p>>=1;
        }
        return exponent<0? 1/r: r;
    }
};