Question

我正在为Android制作应用程序。我试图猜测用户刚刚在木琴上播放的是什么歌。我有：private List<String> mDespacito = new ArrayList<String>(Arrays.asList("f", "a","d","d"));和private List<String> mPlayed = new ArrayList<String>();当用户按下木琴上的一个键时，我将他按下的键添加到mPlayed arraylist中，如下所示：

public void playD(View v){
    Log.d("Xylophone", "Played D!");
  mSoundPool.play(mDSoundId,LEFT_VOLUME,RIGHT_VOLUME,PRIORITY,NO_LOOP,NORMAL_PLAY_RATE);
    mPlayed.add("d");
    CheckForSong();

}

现在，CheckForSong包含：

public void CheckForSong(){
    if (mPlayed.containsAll(mDespacito)){
        Log.d("Xylophone","You just played despacito");
        mPlayed.removeAll(mDespacito);
    }
}

所以，应该这样做：

played F
played A
played D
played D
You just played despacito

但确实如此：

played F
played A
played D
You just played despacito
played D

你甚至可以做到：

played F
played G
played A
played G
played D
You just played despacito

我知道原因：因为if (mPlayed.containsAll(mDespacito))只检查mDespacito的元素是否在mPlayed中。但我需要检查是否有mDespacito的所有元素（包括那些有两次的元素）以及它们是否按照正确的顺序排列。有没有我可以使用的命令？谢谢

Answer 1

使用

{
    "$schema": "http://json-schema.org/draft-06/schema#",
    "title": "Product set",
    "type": "array",
    "items": {
        "title": "Product",
        "type": "object",
        "properties": {
            "id": {
                "description": "The unique identifier for a product",
                "type": "number"
            },
            "name": {
                "type": "string"
            },
            "price": {
                "type": "number",
                "exclusiveMinimum": 0
            },
            "dimensions": {
                "type": "object",
                "properties": {
                    "length": {"type": "number"},
                    "width": {"type": "number"},
                    "height": {"type": "number"}
                },
                "required": ["length", "width", "height"]
            },
            "warehouseLocation": {
                "description": "Coordinates of the warehouse with the product",
                "$ref": "http://json-schema.org/geo"
            }
        },
        "required": ["id", "name", "price"]
    }
}

而是

，因此将按顺序和元素检查元素。

重要提示：如果您没有使用Strings作为代码演示，则需要在添加到列表中的类中实现hashCode和equals。

以下代码段导致显示true两次，然后显示false

mPlayed.equals(mDespacito);

另外：您可以自己比较一下这个列表：

import java.util.ArrayList;

public class MyClass {
    public static void main(String args[]) {
        ArrayList<String> a = new ArrayList();
        a.add("f");
        a.add("a");
        a.add("d");
        a.add("d");

        ArrayList<String> b = new ArrayList();
        b.add("f");
        b.add("a");
        b.add("d");
        b.add("d");

        System.out.println(a.equals(b));
        System.out.println(b.equals(a));
        b.add("c");
        System.out.println(a.equals(b));
    }
}

但请记住，如果您不使用基元或String，则需要在对象上实现hashCode和equals。

Answer 2

Collections.indexOfSubList就是答案。我这样用它：

public void CheckForSong(){
    int contains = Collections.indexOfSubList(mPlayed, mDespacito);
    if (contains != -1){
        Log.d("Xylophone","You just played despacito");
        mPlayed.removeAll(mDespacito);
    }
}

有关Collections.indexOfSubList的更多信息： https://www.tutorialspoint.com/java/util/collections_indexofsublist.htm

Answer 3

list1.equals(list2)可以使用

以下链接给出了一个非常好的解释。请找。

https://stackoverflow.com/a/1075699/4994582

Java是

3 个答案: