建立基于多个参数的排序列表？

Question

我有三个数组

String[] persons = {"jack","james","hill","catnis","alphonso","aruba"};
int[] points = {1,1,2,3,4,5};
int[] money = {25,66,24,20,21,22};

这三个数组中的第n个位置属于同一个实体，例如：-

个人==指出==的钱，即杰克有1分和25美元的。

我想要构建一个对person alphabetically(ascending)排序的列表，如果起始字母相同，那么它应该检查points(descending)，如果它们也是相同的，那么它必须检查money(descending).

排序后的最终列表应该是{阿鲁巴、阿方索、卡尼斯、希尔、詹姆斯、杰克}。

Answer 1

所以我想你想要这样的东西：

public class Person {
   String name;
   int points;
   int money;

   public Person(String name, int points, int money) {
       this.name = name;
       this.points = points;
       this.money = money;
   }

   // getters
}

然后用您拥有的数据(例如，List<Person> )创建一个new Person("jack", 1, 25)。然后分类：

Collections.sort(persons, (person1, person2) -> {
    // could be written more concisely, but this should make things clear
    char letter1 = person1.getName().charAt(0);
    char letter2 = person2.getName().charAt(0);
    if (letter1 != letter2) {
        return letter1 - letter2;
    }
    int points1 = person1.getPoints();
    int points2 = person2.getPoints();
    if (points1 != points2) {
        return points2 - points1; // notice the order is reversed here
    }
    int money1 = person1.getMoney();
    int money2 = person2.getMoney();
    if (money1 != money2) {
        return money2 - money1;
    }
    return 0; // unless you want to do something fancy for tie-breaking
});

这将根据您的标准给出一个排序的List<Person>。

Answer 2

如果你想干些又快又脏的事：

    Comparator<Integer> cName = (i, j) -> Character.compare( persons[i].charAt(0), persons[j].charAt(0));
    Comparator<Integer> cPoints = (i, j) -> Integer.compare( points[i], points[j]);
    Comparator<Integer> cMoney = (i, j) -> Integer.compare( money[i], money[j]);

    List<String> l = 
            IntStream.range(0, persons.length).boxed()
            .sorted( cName.thenComparing(cPoints.reversed()).thenComparing(cMoney.reversed()) )
            .map( i -> persons[i] )
            .collect(Collectors.toList());

    System.out.println(l);

前3行使用lambda定义基于数组索引的比较器。

下面的行使用流：

1. 创建索引的int流，从0到Persons.ength-1
2. 基于比较器序列的流排序索引
3. 将排序索引映射到人名
4. 把它整理成一张清单

兰达和溪流不是很酷吗？

Answer 3

如果您可以拥有一个Person模型：

final class Person {
  private final String name;

  private final int points;

  private final int money;

  public Person(final String name, final int points, final int money) {
    this.name = name;
    this.points = points;
    this.money = money;
  }

  // getters and setters (if you want)

  @Override
  public String toString() {
    final StringBuffer sb = new StringBuffer("Person {")
        .append("name=")
        .append(name)
        .append(", points=")
        .append(points)
        .append(", money=")
        .append(money)
        .append('}');
    return sb.toString();
  }
}

然后你可以这样做：

public static void main(final String... args) throws Exception {
  Person[] persons = new Person[6]; // Use a List (if you can)
  persons[0] = new Person("jack", 1, 25);
  persons[1] = new Person("james", 1, 66);
  persons[2] = new Person("hill", 2, 24);
  persons[3] = new Person("catnis", 3, 20);
  persons[4] = new Person("alphonso", 4, 21);
  persons[5] = new Person("aruba", 5, 22);
  System.out.printf("persons = %s%n%n", Arrays.toString(persons));
  System.out.printf("Person[0] = %s%n%n", persons[0]);
  Collections.sort(Arrays.asList(persons), (c1, c2) -> {
    final int charComp = Character.compare(c1.name.charAt(0), c2.name.charAt(0));
    if (0 == charComp) {
      final int pointsComp = Integer.compare(c2.points, c1.points);
      if (0 == pointsComp) { return Integer.compare(c2.money, c1.money); }
      return pointsComp;
    }
    return charComp;
  });
  // The collection was modified at this point because of the "sort"
  System.out.printf("persons = %s%n", Arrays.toString(persons));
}

结果：

Person = Person {name=jack，points=1，money=25}，Person {name=james，points=1，money=66}，Person {name=hill，points=2，money=24}，Person {name=catnis，points=3，money=20}，Person {name=alphonso，points=4，money=21}，Person {name=aruba，points=5，money=21} Person = Person {name=jack，points=1，money=25} Person = Person {name=aruba，points=5，money=22}，Person {name=alphonso，points=4，money=21}，Person {name=catnis，points=3，money=20}，Person {name=hill，points=2，money=24}，Person {name=james，points=1，money=66}，Person {name=jack，points=1，money=66}

一个更紧凑的sort (但效率有点低，因为您必须预先运行所有比较)：

Collections.sort(Arrays.asList(persons), (c1, c2) -> {
  final int names = Character.compare(c1.name.charAt(0), c2.name.charAt(0));
  final int points = Integer.compare(c2.points, c1.points);
  final int money = Integer.compare(c2.money, c1.money);
  return (0 == names) ? ((0 == points) ? money : points) : names;
});