pow 的实现

要点：

1 n的正负数

2 double 类型的

class Solution {
public:
    double pow(double x, int n) {
        
        if(equal(x,0.0)&&n<0){
            return 0;
        }
        double res=1;
        
        if(n<0){
            res=1.0/power(x,-n);
        }else{
            res=power(x,n);
        }
        
        return res;
            
    }
    double power(double x,int n){
        if(n==0){
            return 1;
        }
        double temp=1;
        if(n%2==0){
            temp*=power(x*x,n/2);    
        }else{
            temp=temp*x*pow(x*x,n/2);
        }
        
        return temp;
    }
    
    bool equal(double num1,double num2){
        if(fabs(num1-num2)<0.000000001){
            return 1;
        }
        return 0;
    }
};