数据结构按值排序

Question

I am trying to store a score for a large about of players. So I will need a map that is sorted based on the value, but because of the amount of players it's inefficient to sort it each time I retrieve it. I would also like the ability to find a players rank in the map. It would be very similar to the score datatype in redis. Something like:

    ScoreMap<String, Integer> scores = new ScoreMap<String, Integer>();

    scores.put("Bill", 2);
    scores.put("Tom", 6);
    scores.put("Jim", 3);
    scores.put("Jake", 3);


    System.out.println("Rank = " + scores.getRank("Bill"));

    System.out.println();
    System.out.println("All:");
    for (Entry<String, Integer> entry : scores.entrySet()) {
        System.out.println(entry.getKey() + " => " + entry.getValue());
    }

    System.out.println();
    System.out.println("Rank Range:");
    for (Entry<String, Integer> entry : scores.entryRankRange(0, 2)) {
        System.out.println(entry.getKey() + " => " + entry.getValue());
    }

    System.out.println();
    System.out.println("Score Range:");
    for (Entry<String, Integer> entry : scores.entryScoreRange(2, 3)) {
        System.out.println(entry.getKey() + " => " + entry.getValue());
    }

This would return

    Rank = 3

    All:
    Tom => 6
    Jake => 3
    Jim => 3
    Bill => 2

    Rank Range:
    Tom => 6
    Jake => 3
    Jim => 3

    Score Range:
    Jake => 3
    Jim => 3
    Bill => 2

I know this is kinda specific and I will probably have to make a custom data-structure. But a point in the right direction would be much appreciated. :)

Answer 1

Simplest way to do this would be to use a Set (namely, TreeSet ) and encapsulate the player information (including the score) into a specific class:

public class CompetitivePlayer implements Comparable<CompetitivePlayer>{

    private String name;
    private int score;    

    public CompetitivePlayer(String name, int score) {
        this.name = name;
        this.score = score;
    }

    public String getName() {
        return name;
    }

    public int getScore() {
        return score;
    }

    public void incrementScore() {
        score++;
    }

    @Override
    public int compareTo(CompetitivePlayer o) {
        return score - o.score;
    }
}

TreeSet assumes that entires stored in it implement the Comparable interface for determining the natural ordering of its elements.

Edit:

If you need to frequently modify the scores of players, then a Map<String, Integer> -based solution is a better fit, because there's no get in Java's Set . This thread discusses the Map -based approach in great detail.

One simplictic solution (as suggested in the mentioned thread) is a one-liner, using the Guava library:

Map<String, Integer> sortedScores = ImmutableSortedMap.copyOf(scores,
    Ordering.natural().onResultOf(Functions.forMap(scores)));

Answer 2

Use Collections .sort() and provide a custom Comparator to sort a map based on its values :

Collections.sort( map, new Comparator<Map.Entry<K, V>>()
{
    @Override
    public int compare( Map.Entry<K, V> entry1, Map.Entry<K, V> entry2)
    {
        return (entry1.getValue()).compareTo( entry2.getValue() );
    }
} );

Answer 3

Can you rewrite your name and score as a class player, then you can use TreeSet to keep the order, and override equals() and compareTo(), it seems like you want to use name as identifier, and keep players ordered by score, then you can do

@Override
public boolean equals(Object obj) {
    if(obj instanceof Player)
        return this.name.equals(((Player)obj).name);
    return false;
}
@Override
public int compareTo(Player p) {
    return p.score - this.score;
}