【ATT】 Implement pow(x, n)

    double helper(double x, unsigned int n)
    {
        if(n==1)
          return x;
        double result;
        if(n%2==0)
        {
            result = helper(x,n/2);
            return result*result;
        }else{
            result = helper(x,(n-1)/2);
            return result*result*x;
        }
          
        
    }
    double pow(double x, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n==0)
          return 1.0;
          
        bool bNegative = (n<0)?true:false;
          
        if(bNegative)
            return 1.0/helper(x,abs(n));
        else 
            return helper(x,abs(n));  
    }