用多个if-else语句（多个条件）优化函数的最佳方法是什么？

Question

什么是优化下面代码的最佳方法或设计模式是什么？（我曾考虑过使用switch语句，但是switch语句无法在单个情况下处理多个条件。）

下面是代码片段。 每个专业都由某个数值范围确定。

public String getMajor(String major) {
    crnCompare = Integer.parseInt(major);
    if ((crnCompare >= 90702 && crnCompare <= 90733) || (crnCompare >= 10004 && crnCompare <= 10037)) {
        this.major = "AC";
    } else if ((crnCompare >= 10087 && crnCompare <= 10108) || (crnCompare >= 10471 && crnCompare <= 10482) || (crnCompare >= 90024 && crnCompare <= 90071)) {
        this.major = "CS";
    } else if ((crnCompare >= 10109 && crnCompare <= 10158) || (crnCompare >= 90072 && crnCompare <= 90116)) {
        this.major = "EC";
    } else if ((crnCompare >= 90117 && crnCompare <= 90203) || (crnCompare >= 10075 && crnCompare <= 10213) || (crnCompare >= 10498 && crnCompare <= 10572)) {
        this.major = "EN";
    } else if ((crnCompare >= 10038 && crnCompare <= 10040) || (crnCompare >= 10214 && crnCompare <= 10255) || (crnCompare >= 10256 && crnCompare <= 10260) || (crnCompare >= 90017 && crnCompare <= 90203) || crnCompare == 11172) {
        this.major = "FI";
    } else if ((crnCompare >= 90670 && crnCompare <= 90790) || (crnCompare >= 11236 && crnCompare <= 11239)) {
        this.major = "FS";
    } else if ((crnCompare >= 90253 && crnCompare <= 90273) || (crnCompare >= 90734 && crnCompare <= 90769) || (crnCompare >= 90274 && crnCompare <= 90360) || (crnCompare >= 10261 && crnCompare <= 10393)) {
        this.major = "GB";
    } else if ((crnCompare >= 100394 && crnCompare <= 10429) || (crnCompare >= 90361 && crnCompare <= 90398)) {
        this.major = "GLS";
    } else if ((crnCompare >= 10430 && crnCompare <= 10451) || (crnCompare >= 90399 && crnCompare <= 90420)) {
        this.major = "HI";
    } else if ((crnCompare >= 10452 && crnCompare <= 10468) || (crnCompare >= 90422 && crnCompare <= 90436) || crnCompare == 11119) {
        this.major = "IDCC";
    } else if ((crnCompare >= 9437 && crnCompare <= 90438) || (crnCompare >= 10469 && crnCompare <= 10470)) {
        this.major = "IPM";
    } else if ((crnCompare == 90421) || (crnCompare >= 11280 && crnCompare <= 11426)) {
        this.major = "ID";
    } else if ((crnCompare >= 90439 && crnCompare <= 90448) || (crnCompare >= 90483 && crnCompare <= 90497)) {
        this.major = "LTF";
    } else if ((crnCompare >= 90504 && crnCompare <= 90535) || (crnCompare >= 10573 && crnCompare <= 10596) || crnCompare == 90785) {
        this.major = "MG";
    } else if ((crnCompare >= 90536 && crnCompare <= 90553) || (crnCompare >= 10598 && crnCompare <= 10616) || crnCompare == 10740) {
        this.major = "MK";
    } else if ((crnCompare >= 90449 && crnCompare <= 90503) || (crnCompare >= 10514 && crnCompare <= 10564) || (crnCompare == 11120) || (crnCompare == 10555) || (crnCompare == 11127)) {
        this.major = "MA";
    } else if ((crnCompare >= 10637 && crnCompare <= 10715) || (crnCompare == 11142) || (crnCompare == 10739) || (crnCompare >= 90575 && crnCompare <= 90622)) {
        this.major = "NAS";
    } else if (crnCompare >= 90554 && crnCompare <= 90574 || crnCompare == 10617 || crnCompare == 10636) {
        this.major = "ML";
    } else if ((crnCompare >= 90623 && crnCompare <= 10646) || (crnCompare >= 10671 && crnCompare <= 10696)) {
        this.major = "PI";
    } else if ((crnCompare == 90647 || crnCompare == 90649) || (crnCompare >= 10697 && crnCompare <= 10698) || crnCompare == 10756) {
        this.major = "PRS";
    } else if ((crnCompare >= 11341 && crnCompare <= 11420)) {
        this.major = "SL";
    } else if ((crnCompare >= 90650 && crnCompare <= 90668) || (crnCompare >= 10716 && crnCompare <= 10734)) {
        this.major = "SO";
    } else if ((crnCompare == 10735)) {
        this.major = "ST";
    }

    return this.major;
}

Answer 1

您可能想研究Guava的RangeMap类（其他类似的实现方式也可用）。

这些使您可以表达这些条件，如下所示：

RangeMap<Integer, String> rangeMap =
  ImmutableRangeMap.<Integer, String>builder()
    .put(Range.closed(90702, 90733), "AC")
    .put(Range.closed(10004, 10037), "AC")
    .put(Range.closed(10087, 10108), "EN")
    .put(Range.closed(10004, 10037), "AC")
    // ...
    .build();

构造一次，然后查询如下：

String major = rangeMap.get(crmCompare);

有两个优点：

• 这是一种更紧凑的语法
• 验证了范围的创建，最小和最大均处于正确的顺序，并且没有重叠。

缺点是添加了番石榴，如果您还没有使用过的话。

Answer 2

标准表驱动方法：

public static class Range
{
    // getters omitted for conciseness
    int low;
    int high;
    String major;
    public Range(int low, int high, String major)
    {
        this.low  = low;
        this.high  = high;
        this.major = major;
    }
    public boolean contains(int v)
    {
        return (v >= low && v <= high);
    }
}

public static Range[] ranges = {
        new Range(10004,10037,"AC"),
        new Range(10087,10108,"AC"),
        // etc
        // Ideally this table is populated from a data file that can
        // be updated at runtime without recompiling the code.
};

public String getMajor(String m)
{
    int crnCompare = Integer.parseInt(m);
    // Search for the matching range
    for (Range r : ranges)
        if (r.contains(crnCompare)) return r.major;
    return null;
}

将所有条件编码为低-高对和相应的主代码，然后将所有条件放入数组中。 要确定主要搜索数组，以找到匹配条件。

可能的增强（保留为练习）包括

1. 从文件加载表，而不是对其进行硬编码
2. 保持表格排序，以便您可以执行二进制搜索而不是线性搜索。

Answer 3

我想了一会儿。 关注以下方面：“如何确保实现正确？”

从这个意义上讲，我会建议一个Range类，类似于Jim提出的类。 因为这样的类使您不仅可以拥有一个与范围数据紧密相关的contains()方法，而且还可以使您拥有一个与范围数据紧密相关的contains()方法。 但最重要的是，您可以添加以下方法：

• boolean isOverlapping(Range other) ...可以用来检测不注意的情况，并且您定义的范围可以重叠！
• int compareTo(Range other) ...实现Comparable接口。 现在您可以对范围进行排序 。 当某个“类别”的范围列表太大时，这将允许您执行二进制搜索。

但是：我不会将“主要”名称放到Range类中。 但是反过来：

public enum RangeBasedMajor {
  AC(Arrays.asList(new Range(90702, 90733), new Range(...)),
  EN(...

  private RangeBasedMajor(List<Range> ranges) ...

这里的重点：

• 您在一个地方创建了这些“专业”并将其映射到其范围
• 您可以编写一些不错的单元测试，以从该枚举中获取所有常量，然后grep其范围，以确保没有范围重叠
• 而且您仍然可以将所有信息随意放入Andy在其解决方案中推荐给您的特殊地图类型中...

根据您的情况，这可能会过分杀伤力。 这实际上取决于您期望事物在其中发生变化的频率。 例如：如果需要访问/处理程序其他部分中的“主要”信息。 当您只需要将该数字映射到字符串时（例如用于打印内容）； 那么RangeMap就可以了。

但是，当您有其他代码与“专业”打交道时，那么值得在这里提出一些建议。