public static double myPow(double x, int n) {
    if (n == 0) {
        return 1;
    }
    long exp = n;//转化成long防止溢出
    if (exp < 0) {
        exp = -exp;
    }
    double res = recursion(x, exp);

    if (n > 0) {
        return res;
    } else {
        if (res != 0)
            return 1 / res;
        else//这里不确定怎么处理
            return 0;
    }
}

//快速幂
public static double recursion(double x, long n) {
    if (n == 0) return 1;
    if (n == 1) return x;

    double result = recursion(x, n >> 1);

    result = result * result;
    if (n % 2 == 1) result = result * x;
    return result;
}