剑指Offer-13-数值的整数次方

/**
 * 数值的整数次方：https://leetcode-cn.com/problems/shu-zhi-de-zheng-shu-ci-fang-lcof/
 * @Description
 */
public class Test14 {
    boolean isInvalid = false;
    public double myPow(double x, int n) {
        isInvalid = false;
        double result;
        if (n > 0){
            result = powerCore(x,n);
        } else if (n < 0) {
            if (x == 0) {
                isInvalid = true; //0的负数次方不存在
                return 0;
            }
            //x ！= 0
            result = 1 / powerCore(x, -n);
        } else {
            return 1; //这里0的0次方输出为1
        }
        return result;

    }

    private double powerCore(double x, int n) {
        if (n == 1){
            return x;
        }
        if (n == 0){
            return 1;
        }
        double result = powerCore(x,n >> 1);
        result *= result;
        if ((n & 1) == 1){
            result *= x;
        }
        return result;
    }

    //快速幂方法
    public double myPow2(double x, int n) {
        if(x == 0) return 0;
        long b = n;
        double res = 1.0;
        if(b < 0) {
            x = 1 / x;
            b = -b;
        }
        while(b > 0) {
            if((b & 1) == 1) res *= x;
            x *= x;
            b >>= 1;
        }
        return res;
    }

    public static void main(String[] args) {
        System.out.println(new Test14().myPow2(5, -4));
    }
}