java - 使用 if-condition 还是 HashMap？

Question

我在面试中被问到这个问题以改进所提供的代码。提供的代码使用了很多 if 语句，因此我决定使用 HashMap 因为检索会更快。不幸的是，我没有被选中担任该职位。我想知道是否有人知道比我改进代码更好的方法？

/* The following Java code is responsible for creating an HTML "SELECT" list of
   U.S. states, allowing a user to specify his or her state. This might be used,
   for instance, on a credit card transaction screen. 

   Please rewrite this code to be "better". Submit your replacement code, and 
   please also submit a few brief comments explaining why you think your code 
   is better than the sample.  (For brevity, this sample works for only 5 
   states. The real version would need to work for all 50 states. But it is 
   fine if your rewrite shows only the 5 states here.)
 */

/* Generates an HTML select list that can be used to select a specific U.S. 
   state.
 */

public class StateUtils {

  public static String createStateSelectList() {

    return
      "<select name=\"state\">\n"
    + "<option value=\"Alabama\">Alabama</option>\n"
    + "<option value=\"Alaska\">Alaska</option>\n"
    + "<option value=\"Arizona\">Arizona</option>\n"
    + "<option value=\"Arkansas\">Arkansas</option>\n"
    + "<option value=\"California\">California</option>\n"
    // more states here
    + "</select>\n"
    ;
  }


  /* Parses the state from an HTML form submission, converting it to the 
     two-letter abbreviation. We need to store the two-letter abbreviation 
     in our database.
   */

  public static String parseSelectedState(String s) {
    if (s.equals("Alabama"))     { return "AL"; }
    if (s.equals("Alaska"))      { return "AK"; }
    if (s.equals("Arizona"))     { return "AZ"; }
    if (s.equals("Arkansas"))    { return "AR"; }
    if (s.equals("California"))  { return "CA"; }
    // more states here
  }

  /* Displays the full name of the state specified by the two-letter code. */

  public static String displayStateFullName(String abbr) {
  {
    if (abbr.equals("AL")) { return "Alabama";    }
    if (abbr.equals("AK")) { return "Alaska";     }
    if (abbr.equals("AZ")) { return "Arizona";    }
    if (abbr.equals("AR")) { return "Arkansas";   }
    if (abbr.equals("CA")) { return "California"; }
    // more states here
  }
}

我的解决方案

/* Replacing the various "if" conditions with Hashmap<key, value> combination 
   will make the look-up in a constant time while using the if condition 
   look-up time will depend on the number of if conditions.
 */

import java.util.HashMap;

public class StateUtils {

  /* Generates an HTML select list that can be used to select a specific U.S. 
     state.
   */
  public static String createStateSelectList() {
    return "<select name=\"state\">\n"
    + "<option value=\"Alabama\">Alabama</option>\n"
    + "<option value=\"Alaska\">Alaska</option>\n"
    + "<option value=\"Arizona\">Arizona</option>\n"
    + "<option value=\"Arkansas\">Arkansas</option>\n"
    + "<option value=\"California\">California</option>\n"
    // more states here
    + "</select>\n";
  }

  /* Parses the state from an HTML form submission, converting it to the 
     two-letter abbreviation. We need to store the two-letter abbreviation 
     in our database.
   */

  public static String parseSelectedState(String s) {
    HashMap<String, String> map = new HashMap<String, String>();
    map.put("Alabama", "AL");
    map.put("Alaska", "AK");
    map.put("Arizona", "AZ");
    map.put("Arkansas", "AR");
    map.put("California", "CA");

    // more states here

    String abbr = map.get(s);
    return abbr;
  }

  /* Displays the full name of the state specified by the two-letter code. */

  public static String displayStateFullName(String abbr) {
    {

      HashMap<String, String> map2 = new HashMap<String, String>();
      map2.put("AL", "Alabama");
      map2.put("AK", "Alaska");
      map2.put("AZ", "Arizona");
      map2.put("AR", "Arkansas");
      map2.put("CA", "California");

      // more state abbreviations here here

      String full_name = map2.get(abbr);
      return full_name;
    }
  }
}

Answer 1

我认为您的代码有很多问题，尤其是为每个方法调用重新创建 Map。

我会从头开始，从接口(interface)开始。我们需要两件事；一个 State 和一个 StateResolver。界面看起来像这样:

public interface State {

    String fullName();

    String shortName();
}

public interface StateResolver {

    State fromFullName(final String fullName);

    State fromShortName(final String shortName);

    Set<? extends State> getAllStates();
}

这允许在稍后阶段将实现换成更明智的东西，比如数据库。但是让我们坚持使用示例中的硬编码状态。

我会将 State 实现为 enum，如下所示:

public enum StateData implements State {

    ALABAMA("Alabama", "AL"),
    ALASKA("Alaska", "AK"),
    ARIZONA("Arizona", "AZ"),
    ARKANSAS("Arkansas", "AR"),
    CALIFORNIA("Californiaa", "CA");

    private final String shortName;
    private final String fullName;

    private StateData(final String shortName, final String fullName) {
        this.shortName = shortName;
        this.fullName = fullName;
    }

    @Override
    public String fullName() {
        return fullName;
    }

    @Override
    public String shortName() {
        return shortName;
    }
}

但是，如上所述，这可以用从数据库加载的 bean 代替。实现应该是不言自明的。

接下来是解析器，让我们针对我们的 enum 写一个:

public final class EnumStateResolver implements StateResolver {

    private final Set<? extends State> states;
    private final Map<String, State> shortNameSearch;
    private final Map<String, State> longNameSearch;

    {
        states = Collections.unmodifiableSet(EnumSet.allOf(StateData.class));
        shortNameSearch = new HashMap<>();
        longNameSearch = new HashMap<>();
        for (final State state : StateData.values()) {
            shortNameSearch.put(state.shortName(), state);
            longNameSearch.put(state.fullName(), state);
        }
    }

    @Override
    public State fromFullName(final String fullName) {
        final State s = longNameSearch.get(fullName);
        if (s == null) {
            throw new IllegalArgumentException("Invalid state full name " + fullName);
        }
        return s;
    }

    @Override
    public State fromShortName(final String shortName) {
        final State s = shortNameSearch.get(shortName);
        if (s == null) {
            throw new IllegalArgumentException("Invalid state short name " + shortName);
        }
        return s;
    }

    @Override
    public Set<? extends State> getAllStates() {
        return states;
    }

}

同样，这是不言自明的。变量处于实例级别。对 StateData 类的唯一依赖是在初始化程序 block 中。这显然需要为另一个 State 实现重写，但这应该没什么大不了的。请注意，如果状态无效，此类将抛出 IllegalArgumentException - 这需要在某处以某种方式进行处理。目前还不清楚这会发生在哪里，但需要考虑一些事情。

最后我们在类中实现了需要的方法

public final class StateUtils {

    private static final StateResolver STATE_RESOLVER = new EnumStateResolver();
    private static final String OPTION_FORMAT = "<option value=\"%1$s\">%1$s</option>\n";

    public static String createStateSelectList() {
        final StringBuilder sb = new StringBuilder();
        sb.append("<select name=\"state\">\n");
        for (final State s : STATE_RESOLVER.getAllStates()) {
            sb.append(String.format(OPTION_FORMAT, s.fullName()));
        }
        sb.append("</select>\n");
        return sb.toString();
    }

    public static String parseSelectedState(final String s) {
        return STATE_RESOLVER.fromFullName(s).shortName();
    }

    public static String displayStateFullName(final String abbr) {
        return STATE_RESOLVER.fromShortName(abbr).fullName();
    }
}

请注意，我们仅在实用程序类的顶部引用了实现，这样可以快速轻松地换出实现。我们使用 static final 引用来表示 StateResolver 只创建一次。我还将选择的硬编码创建替换为基于动态循环的选择。我还使用格式化程序来构建选择。

应该注意的是，用 Java 构建 HTML绝不是一个好主意，任何这样做的人都应该对他们做一些难以形容的事情。

不用说，您应该对上述代码的每一行进行彻底的单元测试。

简而言之，您的答案并没有真正接近针对手头问题的适当、可扩展的企业解决方案。我的解决方案可能看起来有些矫枉过正，而你可能是对的。但我认为这是正确的方法，因为抽象是可重用代码的关键。