剑指offer——数值的整数次方

/**
* 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。
*@author: Administrator
*@date: 2017-1-15 下午04:21:41
*/
import java.util.Scanner;
public class Solution {
public double Power(double base, int exponent) {

if(base==0&&exponent<0)return 0;

if(exponent==0)return 1;
if(exponent==1)return base;

int abs=Math.abs(exponent);
double result=Power(base,abs>>1);
result*=result;
if((abs&0x1)==1)result*=base;

if(exponent<0)return 1/result;

return result;
}
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int exponent=sc.nextInt();
double base=sc.nextDouble();
sc.close();
Solution s=new Solution();
System.out.println(s.Power(base, exponent));
}

}