java - 这是按 Java 集合中的键排序的最有效方法吗？

Question

有没有更有效的方法来实现这个结果。我本质上想先看看值Section最长的getContextPath()。

enum Section {
    Provider("Provider", "com.app.provider"),
    Container("Container", "com.app");

    /**
     * The key for this Section.
     */
    private String key;

    /**
     * The context path for this Section.
     */
    private String contextPath;

    /**
     * Create a new Section.
     *
     * @param key         The Key for this section.
     * @param contextPath The Context Path for this section.
     */
    Section(String key, String contextPath) {
        this.key = key;
        this.contextPath = contextPath;
    }

    /**
     * Retrieve the String representation of this Section.
     *
     * @return The String value.
     */
    @SuppressWarnings("NullableProblems")
    @Override
    public String toString() {
        return this.key;
    }

    /**
     * Retrieve the Context Path for this Section.
     *
     * @return The Context Path.
     */
    public String getContextPath()
    {
        return this.contextPath;
    }
}

// Keep in mind that these may be dynamically loaded, and won't always be
// statically loaded like this.
private static final Map<Section, String> sectionLoggerIds = new HashMap<>() {{ 
    put(Section.Container, "some.id");
    put(Section.Provider, "some.other.id");
}};

/**
 * Retrieve a Logger based on a Context.
 *
 * @param context The Context.
 * @return The Logger
 */
public static Logger logger(Context context)  {
    // Create a TreeSet for sorting the values, and set it up
    // to sort them longest first.
    Set<Section> sectionSet = new TreeSet<>(new Comparator<Section>() {
        @Override
        public int compare(Section section1, Section section2) {
            return Integer.compare(
                section2.getContextPath().length(), 
                section1.getContextPath().length()
            );
        }
    });

    // Add all of the relevant Section objects.
    sectionSet.addAll(sectionLoggerIds.keySet());

    // Search the Sections, starting with the longest
    // contextPath first.
    for(Section section : sectionSet) {
        if (context.getClass().getName().contains(section.getContextPath())) {
            return getLogger(sectionLoggerIds.get(section));
        }
    }

    return null;
}

标签： javaandroidsortingset