快速幂板子

public class Solution {
    public double Power(double base, int exponent) {
        if(base == 0)return 0;
        else if(exponent == 0)return 1;
        if(exponent > 0)return quickPower(base, exponent);
        else return 1.0 / quickPower(base, -exponent);
  }

    private double quickPower(double base, int exponent){
        double res = 1;
        while(exponent > 0){
            if((exponent & 1) == 1)res *= base;
            base = base * base;
            exponent >>= 1;
        }
        return res;
    }
}