快速幂解决数值的整数次方

public class Solution {
    public double Power(double base, int exponent) {
        boolean flag = false;
        if(exponent < 0){
            flag = true;
            exponent = -exponent;
        }
        double res = 1;
        while(exponent > 0){
            if((exponent & 1) == 1){
                res = res * base;
            }
            exponent >>= 1;
            base *= base;
        }
        return flag ? 1 / res : res;
  }
}