2个Double字符串进行

public static int compare(double d1, double d2) {
        if (d1 < d2)
            return -1;           // Neither val is NaN, thisVal is smaller
        if (d1 > d2)
            return 1;            // Neither val is NaN, thisVal is larger

        // Cannot use doubleToRawLongBits because of possibility of NaNs.
        long thisBits    = Double.doubleToLongBits(d1);
        long anotherBits = Double.doubleToLongBits(d2);

        return (thisBits == anotherBits ?  0 : // Values are equal
                (thisBits < anotherBits ? -1 : // (-0.0, 0.0) or (!NaN, NaN)
                 1));                          // (0.0, -0.0) or (NaN, !NaN)
    }

　　Double底层实现，比较的时候不能直接用==逻辑运算符来进行判断，因为会存在精度问题，所以转为long类型进行判断更为合适