Java中if的判断条件很多怎么简化

Question

现在有一个需求是如果是某个商户下的设备，就做相应的处理，设备有40个怎么简化。

if(number == "num1" || number == "num2"|| number == "num3"|| number == "num4"|| number == "num5"|| number == "num6"|| number == "num7"|| number == "num8"|| number == "num9"|| number == "num10"|| number == "num11"|| number == "num12"|| number == "num13"|| number == "num14"|| number == "num15"|| number == "num16"|| number == "num17"|| number == "num18"|| number == "num19"|| number == "num20"|| number == "num21|| number == "num23"|| number == "num24"|| number == "num25"|| number == "num26...." ){
    //做相应处理
}

类似这样的，有没有简单点的办法呢

谢谢大佬门啦

找到解决办法啦，谢谢大家。
下面贴出来

        List<String> list = new ArrayList<>();
        list.add("XB203039");
        list.add("XB203031");
                ......


        if(list.contains("XB20303")){
            System.out.println("true");
            ......
        }

Answer 1

Boolean b = false;
for (int i = 1; i < 27; i++)
{
    if (number.equals("num" + i)) { b= true; break;}
}
if (b)
...