Question

在Something like 'contains any' for Java set?中有几种解决方案

Collections.disjoint（A，B）
setA.stream（）。anyMatch（setB :: contains）
Sets.intersection（set1，set2）.isEmpty（）
CollectionUtils.containsAny（）

我的情况是set1是new ConcurrentHashMap<>().keySet()，set2是ArrayList

set1最多可以包含100个条目，set2少于10个条目

还是会全部一样做并且表现相似？

Answer 1

public static void main(String[] args) {
    Map<String, String> map = new ConcurrentHashMap<>();
    List<String> list = new ArrayList<>();

    for (int i = 0; i < 100; i++) {
        map.put(RandomStringUtils.randomNumeric(5), RandomStringUtils.randomNumeric(5));
    }

    for (int i = 0; i < 10; i++) {
        list.add(RandomStringUtils.randomNumeric(5));
    }

    Set<String> set = new HashSet<>(list);

    List<Runnable> methods = new ArrayList<>();
    methods.add(() -> { Collections.disjoint(map.keySet(), list); });
    methods.add(() -> { Collections.disjoint(list, map.keySet()); });

    methods.add(() -> { map.keySet().stream().anyMatch(list::contains); });
    methods.add(() -> { list.stream().anyMatch(map.keySet()::contains); });

    methods.add(() -> { Sets.intersection(map.keySet(), set).isEmpty(); });
    methods.add(() -> { Sets.intersection(set, map.keySet()).isEmpty(); });

    methods.add(() -> { CollectionUtils.containsAny(map.keySet(), list); });
    methods.add(() -> { CollectionUtils.containsAny(list, map.keySet()); });

    for (Runnable method : methods) {
        long start = System.currentTimeMillis();
        for (int i = 0; i < 100000; i++) {
            method.run();
        }
        long end = System.currentTimeMillis();
        System.out.println("took " + (end - start));
    }

}

获奖者iiis Collections.disjoint

took 15
took 32
took 484
took 62
took 157
took 47
took 24
took 32

Answer 2

setA.stream().anyMatch(setB::contains)最好，因为所有其他选项都是非延迟评估，并且将在所有元素上执行。

对于流，它将是惰性计算，一旦找到任何匹配项，就会返回。

另外，来自Documentation of CollectionUtils.containsAny()

换句话说，如果coll1和coll2的交集（java.lang.Iterable，java.lang.Iterable）不为空，则此方法返回true。

进行containsAny检查的最快方法是什么？

2 个答案: