Question

假设您有Arraylist个HockeyPlayer个对象。

如果他们都有变量 int goalsScored ，你如何排序？你怎么能按目标排序呢？

Answer 1

您可以将Collections.sort与自定义Comparator<HockeyPlayer>一起使用。

    class HockeyPlayer {
        public final int goalsScored;
        // ...
    };

    List<HockeyPlayer> players = // ...

    Collections.sort(players, new Comparator<HockeyPlayer>() {
        @Override public int compare(HockeyPlayer p1, HockeyPlayer p2) {
            return p1.goalsScored - p2.goalsScored; // Ascending
        }

    });

比较部分也可以这样写：

players.sort(Comparator.comparingInt(HockeyPLayer::goalsScored));

或者，您可以HockeyPlayer implements Comparable<HockeyPlayer>。这为所有HockeyPlayer个对象定义了自然顺序。使用Comparator更灵活，因为不同的实现可以按名称，年龄等排序。

另见

Java: What is the difference between implementing Comparable and Comparator?

为了完整起见，我应该提醒一下，由于可能出现溢出，必须非常谨慎地使用return o1.f - o2.f逐个减法比较快捷方式（阅读： Effective Java 2nd Edition：第12项：考虑实施{{ 1}} 的）。据推测，曲棍球并不是一项运动，玩家可以在可能导致问题的数量上进球。=）

另见

Java Integer: what is faster comparison or subtraction?

Answer 2

正如@ user6158055所说，它是Java 8的一个班轮，如下所示：

Collections.sort(
                hockeyPlayerList,
                (player1, player2) -> player1.getGoalsScored()
                        - player2.getGoalsScored());

描述相同的完整示例：

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Main {

    public static void main(String[] args) {
        List<HockeyPlayer> hockeyPlayerList = new ArrayList<>();
        hockeyPlayerList.add(new HockeyPlayer("A", 3));
        hockeyPlayerList.add(new HockeyPlayer("D", 10));
        hockeyPlayerList.add(new HockeyPlayer("B", 2));

        System.out.println("Before Sort based on goalsScored\n");

        hockeyPlayerList.forEach(System.out::println);

        System.out.println("\nAfter Sort based on goalsScored\n");

        Collections.sort(
                hockeyPlayerList,
                (player1, player2) -> player1.getGoalsScored()
                        - player2.getGoalsScored());

        hockeyPlayerList.forEach(System.out::println);
    }

    static class HockeyPlayer {

        private String name;
        private int goalsScored;

        public HockeyPlayer(final String name, final int goalsScored) {
            this.name = name;
            this.goalsScored = goalsScored;
        }

        public String getName() {
            return name;
        }

        public void setName(String name) {
            this.name = name;
        }

        public int getGoalsScored() {
            return goalsScored;
        }

        public void setGoalsScored(int goalsScored) {
            this.goalsScored = goalsScored;
        }

        @Override
        public String toString() {
            return "HockeyPlayer [name=" + name + ", goalsScored="
                    + goalsScored + "]";
        }

    }
}

<强>输出：

Before Sort based on goalsScored

HockeyPlayer [name=A, goalsScored=3]
HockeyPlayer [name=D, goalsScored=10]
HockeyPlayer [name=B, goalsScored=2]

After Sort based on goalsScored

HockeyPlayer [name=B, goalsScored=2]
HockeyPlayer [name=A, goalsScored=3]
HockeyPlayer [name=D, goalsScored=10]

Answer 3

编写自定义Comparator来完成工作。

Answer 4

使用类似Bean Comparator的通用比较器。

Answer 5

使用Java 8只需一行：

getComputedStyle()

Answer 6

Java有一组sort（）方法用于此类事情。有关详细信息，请参阅Collections.sort（和Comparable）。

Answer 7

使用Java 8，这很简单

Collections.sort(playList, Comparator.comparingInt(HockeyPLayer::goalsScored))

如何按属性对对象的arylylist进行排序？

7 个答案:

另见

另见