不幸的是，拥有一个按值排序的Map（Map到的对象）不是一件事。相反，您可以有一个键集，它的键按频率排序，但是如果频率是该点的键，那么在事先不知道频率的情况下，就无法在该集中查找条目，这样就消除了练习的重点。
想到的一个策略是有两个独立的数据结构。一种是根据电影名称查找实际对象，另一种是自动排序：

@Data
public class MovieFrequencyTuple implements Comparable<MovieFrequencyTable> {
    @NonNull private final String name;
    private int frequency;

    public void incrementFrequency() {
        frequency++;
    }

    @Override public int compareTo(MovieFrequencyTuple other) {
        int c = Integer.compare(frequency, other.frequency);
        if (c != 0) return -c;
        return name.compareTo(other.name);
    }
}

有了这些：

SortedSet<MovieFrequencyTuple> frequencies = new TreeSet<>();
Map<String, MovieFrequencyTuple> movies = new HashMap<>();

public int increment(String movieName) {
    MovieFrequencyTuple tuple = movies.get(name);
    if (tuple == null) {
        tuple = new MovieFrequencyTuple(name);
        movies.put(name, tuple);
    }

    // Self-sorting data structures will just fail
    // to do the job if you modify a sorting order on
    // an object already in the collection. Thus,
    // we take it out, modify, put it back in.
    frequencies.remove(tuple);
    tuple.incrementFrequency();
    frequencies.add(tuple);
    return tuple.getFrequency();
}

public int get(String movieName) {
    MovieFrequencyTuple tuple = movies.get(movieName);
    if (tuple == null) return 0;
    return tuple.getFrequency();
}

public List<String> getTop10() {
   var out = new ArrayList<String>();
   for (MovieFrequencyTuple tuple : frequencies) {
       out.add(tuple.getName());
       if (out.size() == 10) break;
   }
   return out;
}

每一个操作都要摊销o（1）或o（logn），甚至是前10个操作。因此，如果你运行100万次“增加一部电影的频率，然后获得前10名”，n=#我们这样做的次数，那么最坏的情况是o（nlogn）性能。
注意：将lombok用于构造函数、getter等-如果您不喜欢，请让您的ide生成这些东西。