java - 如何使用流获取多个属性具有最大值的对象？

Question

假设我有一群恶棍。它们的特点是它们有多好、有多坏或有多丑。

static class Villain {
    String name;
    int good;
    int bad;
    int ugly;

    Villain(String name, int good, int bad, int ugly) {
        this.name = name;
        this.good = good;
        this.bad = bad;
        this.ugly = ugly;
    }
}

好吧，认识一下这帮人:

List<Villain> villains = new ArrayList<>();
villains.add(new Villain("Bob", 2, 2, 1));
villains.add(new Villain("Charley", 2, 1, 2));
villains.add(new Villain("Dave", 2, 1, 1));
villains.add(new Villain("Andy", 2, 2, 2));
villains.add(new Villain("Eddy", 1, 2, 2));
villains.add(new Villain("Franz", 1, 2, 1));
villains.add(new Villain("Guy", 1, 1, 2));
villains.add(new Villain("Harry", 1, 1, 1));

我想做的是，我想弄清楚谁是最好的，最坏的，最丑的。我的意思是找出谁是最好的。如果平局，谁是最差的。如果平局，谁最丑。

我已经成功地做到了，代码如下。

List<Villain> bestVillains = villains
        .stream()
        .collect(groupingBy(v -> v.good, TreeMap::new, toList()))
        .lastEntry()
        .getValue()
        .stream()
        .collect(groupingBy(v -> v.bad, TreeMap::new, toList()))
        .lastEntry()
        .getValue()
        .stream()
        .collect(groupingBy(v -> v.ugly, TreeMap::new, toList()))
        .lastEntry()
        .getValue();

这确实会导致 List<Villain>只有一个成员:安迪。他真的是最好的，最坏的，最丑的!

但是，我有相当多的代码重复、收集值、再次将它们转换为流等。关于如何清理它的任何建议？

JVM 是如何处理的。顺序地还是引擎盖下发生了一些魔法？

Answer 1

So the idea is that it first looks at which has the highest value for good, then (in case of a tie) which has the highest value for bad, then (if it is still not decisive) which has the highest value for 'ugly'

您更愿意使用以下 Comparator 对 Villian 进行排序:

Comparator<Villain> villainComparator = Comparator.comparingInt(Villain::getGood)
    .thenComparingInt(Villain::getBad)
    .thenComparingInt(Villain::getUgly);

Villain result = villains.stream()
                         .max(villainComparator)
                         .orElse(null);