如何检查两个流是否不相交? [英] How to check if two streams are disjoint?

问题描述

我想与流进行比较，并检查它们是否具有1个或多个共同的元素(发现1足以停止寻找更多元素).我希望能够将其应用于包含自定义类的Streams.

I would like to compare to streams, and check if they have 1 or more elements in common (finding 1 is sufficient to stop looking for more). I want to be able to apply this to Streams containing a custom-created class.

为说明起见，假设我有一个看起来像这样的课程:

For illustration, let's say I have a class that looks like:

public class Point {
    public final int row;
    public final int col;

    public Point(int row, int col) {
        this.row = row;
        this.col = col;
    }    

    @Override
    public boolean equals(Object obj) {
        if (obj == null) return false;
        if (obj.getClass() != this.getClass()) return false;
        final Point other = (Point) obj;
        return this.row == other.row && this.col == other.col;
    }

    @Override
    public int hashCode() {
        return Objects.hash(row, col); 
    }
}

然后我有两个看起来很可爱的流:

And then I have two lovely streams that look like:

Stream<Point> streamA = Stream.of(new Point(2, 5), new Point(3, 1));
Stream<Point> streamB = Stream.of(new Point(7, 3), new Point(3, 1));

鉴于这些流共有1个点(即Point(3, 1))，所以我希望最终结果为真.

Given that these Streams have 1 Point in common (namely, Point(3, 1)), I would want the final result to be true.

所需的功能可以如下图所示:

The desired functionality can be pictured as:

public static boolean haveSomethingInCommon(Stream<Point> a, Stream<Point> b){
    //Code that compares a and b and returns true if they have at least 1 element in common
}

推荐答案

首先，您必须将Streams转换为Set或List才能避免出现著名的错误:

First of all you have to convert your Streams to a Set or List to not get the famous error:

java.lang.IllegalStateException: stream has already been operated upon or closed

然后您可以使用anyMatch这样:

public static boolean haveSomethingInCommon(Stream<Coord> a, Stream<Coord> b) {
    Set<Coord> setA = a.collect(Collectors.toSet());
    Set<Coord> setB = b.collect(Collectors.toSet());

    return setA.stream().anyMatch(setB::contains);
}

或者您也可以仅将b流转换为Set并使用:

Or you can convert only the b Stream to a Set and use:

public static boolean haveSomethingInCommon(Stream<Coord> a, Stream<Coord> b) {
    Set<Coord> setB = b.collect(Collectors.toSet());
    return a.anyMatch(setB::contains);
}

在您的方法中，我建议使用Set<Coord>而不是Stream<Coord>作为参数.

I would recommend to Set<Coord> instead of Stream<Coord> as param in your method.

public static boolean haveSomethingInCommon(Set<Coord> a, Set<Coord> b) {
    return a.stream().anyMatch(b::contains);
}

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