剑指12 数值的整数次方

题目描述

给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。

保证base和exponent不同时为0

 

class Solution {
public:
    double Power(double base, int exponent) {
    
        if (exponent==0) return 1;
        double ret=pow(Power(base,abs(exponent)/2),2);
        if (abs(exponent)%2==1)
            ret*=base;
        return exponent<0?1/ret:ret;
        
        
    }
};

//简单快速幂
class Solution {
public:
    double Power(double base, int exponent) {
    
        long long p=abs((long long) exponent);
        double r=1.0;
        while (p){
            if (p&1) r*=base;
            base*=base;
            p>>=1;
            
        }
        return exponent <0 ?1/r :r;
    }
};