数值的整数次方

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。

public class OngoingTimesSquare{
    public static double Power(double base,int exponent){
         if(exponent < 0 ){
            exponent = - exponent;
            return 1/solve(base,exponent);
          }
            renturn solve(base,exponent);
    }

       public double solve(double base, int exponent){
        if(exponent==0) return 1;
        if(exponent%2==1){
            return base*solve(base,(exponent-1)/2)*solve(base,(exponent-1)/2);
        }else{
            return solve(base,exponent/2)*solve(base,exponent/2);
        }
    }
}

或者 找到更好的答案，当时脑子想了下差点也用的a的b次方 可以根据b的数值循环一下相乘就好了

  public double Power(double base, int exponent) {
        if(exponent == 0){
            return 1;
        }else if(exponent > 0){
            double num = base;
            for(int i = 1; i < exponent; i++){
                num = num * base;
            }
            return num;
        }else {
            double nums = base;
            int flag = -exponent;
            for(int i = 1; i < flag; i++){
                nums = nums * base;
            }
            return 1/nums;
        }
  }
}