如何使用 Java 8 根据某些条件检查集合中的值是否唯一

How to check values in collection for uniques by some criteria using Java 8

我有以下 class:

public final class InfoWrapper {
    private int count;
    private int itemCode;
    private String name;

    public InfoWrapper() {
    }

    public InfoWrapper(final int count, final int itemCode, final String name) {
        super();
        this.count = count;
        this.itemCode = itemCode;
        this.name = name;
    }

    public int getCount() {
        return count;
    }

    public int getItemCode() {
        return itemCode;
    }

    public String getName() {
        return name;
    }   
}

这是我的测试 class:

public class TestClass {

    public static void main(String [] args) {
        Deque<InfoWrapper> list = new  ArrayDeque<InfoWrapper>();
        list.add(new InfoWrapper(3, 111, "aaa"));
        list.add(new InfoWrapper(7, 112, "bbb"));
        list.add(new InfoWrapper(12, 113, "ccc"));
        list.add(new InfoWrapper(6, 113, "ccc"));
        list.add(new InfoWrapper(8, 113, "ccc"));
        list.add(new InfoWrapper(3, 113, "ccc"));

        System.out.println("Is good enought : " + isGoodEnought(list, 3));
    }

    private static boolean isGoodEnought(final Deque<InfoWrapper> list, final int maxFailsCount) {
        return !list.stream().limit(maxFailsCount).skip(1).anyMatch((res) -> res.getName().equals("ccc"));
    }
}

我的方法 isGoodEnough() 检查前 3 个元素的名称是否不等于 "ccc",但实际上我需要检查名称是否唯一。

private static boolean isGoodEnought(Deque<InfoWrapper> list,
                                     int maxFailsCount) {
    return list.stream()
               .limit(maxFailsCount)
               .map(InfoWrapper::getName)
               .distinct()
               .count() == maxFailsCount;
}

请注意,即使前两个名称相同,这也会遍历 maxFailsCount 个元素。如果 maxFailsCount 可能是一个很大的值,您可能想要使用

之类的东西
private static boolean isGoodEnought(Deque<InfoWrapper> list, int max) {
    Set<String> seenNames = new HashSet<>();
    for (Iterator<InfoWrapper> i = list.iterator(); i.hasNext() && max-- > 0;)
        if (!seenNames.add(i.next().getName()))
            return false;
    return true;
}