按排序顺序存储条目并检索条目周围的条目

Question

Position是排行榜中的位置，1是得分最高的用户

public class EntryTreeMapOption {

private String uid;
private int score;
private int position;

public EntryTreeMapOption(String uid, int score) {

    this.uid = uid;
    this.score = score;

}

public EntryTreeMapOption() {

}

public String getUid() {
    return uid;
}

public void setUid(String uid) {
    this.uid = uid;
}

public int getScore() {
    return score;
}

public void setScore(int score) {
    this.score = score;
}

public int getPosition() {
    return position;
}

public void setPosition(int position) {
    this.position = position;
}

@Override
public int hashCode() {
    final int prime = 31;
    int result = 1;
    result = prime * result + ((uid == null) ? 0 : uid.hashCode());
    return result;
}

@Override
public boolean equals(Object obj) {
    if (this == obj)
        return true;
    if (obj == null)
        return false;
    if (getClass() != obj.getClass())
        return false;
    EntryTreeMapOption other = (EntryTreeMapOption) obj;
    if (uid == null) {
        if (other.uid != null)
            return false;
    } else if (!uid.equals(other.uid))
        return false;
    return true;
}

@Override
public String toString() {
    return "Entry [uid=" + uid + ", score=" + score + ", position=" + position + "]";
}}

GameBoard类有两个方法：

• submitScore(String uid,int score）每个玩家调用这个方法将他的分数提交到游戏板上。每个玩家/用户只有一个条目，因此，如果玩家多次调用此方法，则存储其最新得分
• GetLeaderBoard（字符串uid)

The leader board is :
Entry [uid=user1, score=14, position=1]
Entry [uid=user2, score=8, position=2]
Entry [uid=user3, score=7, position=3]
Entry [uid=user4, score=7, position=3]
Entry [uid=user5, score=4, position=4]
Entry [uid=user6, score=3, position=5]
Entry [uid=user7, score=3, position=5]
Entry [uid=user8, score=1, position=6]

For user5, entries returned should be :
Entry [uid=user3, score=7, position=3]
Entry [uid=user4, score=7, position=3]
Entry [uid=user5, score=4, position=4]
Entry [uid=user6, score=3, position=5]
Entry [uid=user7, score=3, position=5]

For user4, entries returned should be :
Entry [uid=user1, score=14, position=1]
Entry [uid=user2, score=8, position=2]
Entry [uid=user4, score=7, position=3]
Entry [uid=user5, score=4, position=4]
Entry [uid=user6, score=3, position=5]

For user6, entries returned should be :
Entry [uid=user4, score=7, position=3]
Entry [uid=user5, score=4, position=4]
Entry [uid=user6, score=3, position=5]
Entry [uid=user8, score=1, position=6]

For user7, entries returned should be :

Entry [uid=user4, score=7, position=3]
Entry [uid=user5, score=4, position=4]
Entry [uid=user7, score=3, position=5]
Entry [uid=user8, score=1, position=6]

public class GameDefault2 {

    private TreeMap<EntryMapOption, String> leaderBoardEntryUserMap;

    {

        leaderBoardEntryUserMap = new TreeMap<>(Comparator.comparingInt(EntryTreeMapOption::getScore).reversed()
            .thenComparing(EntryTreeMapOption::getUid));
    }

    @Override
    public void submitScore(String uid, int score) {

        EntryMapOption newEntry = new EntryMapOption(uid, score);
        leaderBoardEntryUserMap.put(newEntry, uid);

    }

    @Override
    public List<EntryMapOption> getLeaderBoard(String uid) {

        System.out.println("---------Current leader board---------");
        leaderBoardEntryUserMap.keySet().forEach(System.out::println);

        List<EntryMapOption> userEntryList = leaderBoardEntryUserMap.entrySet().stream()
                .filter(entry -> uid.equalsIgnoreCase(entry.getKey().getUid())).map(Map.Entry::getKey)
                .collect(Collectors.toList());

        if (userEntryList == null || userEntryList.isEmpty())
            return Collections.emptyList();

        // Incomplete and error prone
        EntryMapOption userEntry = userEntryList.get(0);

        List<EntryMapOption> entriesOptionTwo = new ArrayList<>();
        entriesOptionTwo.add(leaderBoardEntryUserMap.higherKey(userEntry));
        entriesOptionTwo.add(userEntry);
        entriesOptionTwo.add(leaderBoardEntryUserMap.lowerKey(userEntry));

        return entriesOptionTwo;

    }

}

上述代码的问题：

• 何时（理想情况下，在submitScore（）期间）以及如何计算“位置”。虽然map.compute（）用于键，但我想知道它是否可以在任何方面提供帮助！
• 检查“//不完整且易出错”注释下面的代码虽然“高键（）”和“低键（）”很有用，但我不确定如何使用它们来选择特定条目下面和上面的固定数量的条目

*****edit-1*******@Holger的修复解决了以下问题

null

Answer 1

您的equals方法基于ID，但您的compare方法基于score，这会引起Javadoc的问题。

请注意，如果要正确实现map接口，则树映射所维护的顺序（与任何排序映射一样）以及无论是否提供显式比较器，都必须与equals一致。（关于一致等于的精确定义，请参见Comparable或Comparator。）这是因为Map接口是根据equals操作定义的，但排序后的Map使用其compareTo（或compare）方法执行所有键比较，因此从排序后的Map的观点来看，该方法认为相等的两个键是相等的。一个排序映射的行为是很好定义的，即使它的排序与Equals不一致；它只是没有遵守Map接口的通用契约。

问题是，compare方法在对不同属性排序时，会将搜索带向错误的方向。

在您的情况下，这可能是一个较少的问题，尽管您可能有两个相等的对象，但没有相同的分数，导致未来的问题。

当项目在树中时更改得分也可能会引起问题，因此每次项目得分更改时您可能必须删除该项目，然后重新添加它们。

该过程的一个工作示例如下

import java.util.Comparator;
import java.util.TreeMap;

public class Test {

  static class Example {
    final int id;
    final int score;

    Example(int id, int score) {
      this.id = id;
      this.score = score;
    }

    @Override
    public boolean equals(Object obj) {
      if (!(obj instanceof Example)) {
        return false;
      }
      final Example other = (Example) obj;
      return other.id == id;
    }

    @Override
    public int hashCode() {
      return id;
    }

    public int getId() {
      return id;
    }

    @Override
    public String toString() {
      return id + " scored " + score;
    }

    public int getScore() {
      return score;
    }
  }

  public static void main(final String... args) {
    Example a = new Example(1, 10);
    Example b = new Example(2, 30);
    Example c = new Example(3, 1);
    Example d = new Example(4, 10);

    TreeMap<Example, Integer> x = new TreeMap<Example, Integer>(Comparator.comparingInt(Example::getScore).thenComparing(Example::getId));

    x.put(a, a.getScore());
    x.put(b, b.getScore());
    x.put(c, c.getScore());
    x.put(d, d.getScore());

    final Example h2 = x.higherKey(a);
    final Example h1 = h2 == null ? null : x.higherKey(h2);

    final Example l1 = x.lowerKey(a);
    final Example l2 = l1 == null ? null : x.lowerKey(l1);

    System.out.println(h1);
    System.out.println(h2);
    System.out.println(a);
    System.out.println(l1);
    System.out.println(l2);
  }
}