【问题标题】:Random number without duplicate when restart activity重启活动时随机数不重复
【发布时间】:2023-03-08 14:22:01
【问题描述】:

我有活动 A、B 和 C。

在活动 B 中,我有随机数生成器和 Collections.shuffle()。 这给了我下一个活动 C 中的随机数。

然后我使用按钮返回活动 B 并再次重新启动整个过程, 但它给了我 C 中的重复项,因为我猜想整个 Collections.shuffle() 都重新启动了。

当我从 B 转到 C 时,我需要得到不重复的随机数,重复这样做。

活动 B:

    public class MainActivity extends Activity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    this.requestWindowFeature(Window.FEATURE_NO_TITLE);
    this.getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN, WindowManager.LayoutParams.FLAG_FULLSCREEN);
    setContentView(R.layout.activity_main);


Button a = (Button) findViewById(R.id.button1); // Here the R.id.button1 is the button from you design
a.setOnClickListener(new View.OnClickListener() {
public void onClick(View arg0) {
    ArrayList<Integer> randomNumber = new ArrayList<Integer>();
    for (int i = 1; i <= 17; ++i) randomNumber.add(i);
    Collections.shuffle(randomNumber);
    int random = randomNumber.get(0); 
Intent intent=new Intent(MainActivity.this,Main2Activity.class);
intent.putExtra("Value",random);
startActivity(intent);
}
    });

Button b = (Button) findViewById(R.id.button2); // Here the R.id.button1 is the button from you design
b.setOnClickListener(new View.OnClickListener() {
public void onClick(View arg0) {
    ArrayList<Integer> randomNumber = new ArrayList<Integer>();
    for (int i = 18; i <= 35; ++i) randomNumber.add(i);
    Collections.shuffle(randomNumber);
    int random = randomNumber.get(0); 
    Intent intent=new Intent(MainActivity.this,Main2Activity.class);
    intent.putExtra("Value",random);
    startActivity(intent);
}
    });

Button c = (Button) findViewById(R.id.button3); // Here the R.id.button1 is the button from you design
c.setOnClickListener(new View.OnClickListener() {
public void onClick(View arg0) {
    ArrayList<Integer> randomNumber = new ArrayList<Integer>();
    for (int i = 36; i <= 50; ++i) randomNumber.add(i);
    Collections.shuffle(randomNumber);
    int random = randomNumber.get(0); 
    Intent intent=new Intent(MainActivity.this,Main2Activity.class);
    intent.putExtra("Value",random);
    startActivity(intent);

}
    });

活动 C:

    public class Main2Activity extends Activity {


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    this.requestWindowFeature(Window.FEATURE_NO_TITLE);
    this.getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN, WindowManager.LayoutParams.FLAG_FULLSCREEN);
    setContentView(R.layout.activity_main2);

   Button back = (Button) findViewById(R.id.buttonback); // Here the R.id.button1 is the button from you design
back.setOnClickListener(new View.OnClickListener() {
public void onClick(View arg0) {
odstevalnik.cancel();
Intent i = new Intent(Main2Activity.this, MainActivity.class);
startActivity(i);
}
    });

    Bundle bundle = getIntent().getExtras();
int random = bundle.getInt("Value");
TextView text = (TextView) findViewById(R.id.textView1);
if (random==1) {
    text.setText("blabla");
    image.setImageResource(R.drawable.img2);
    stopnja.setImageResource(R.drawable.stopnja1);
    Toast.makeText(getApplicationContext(), "blabla", 
    Toast.LENGTH_LONG).show();
}

   .....

【问题讨论】:

  • 你能通过例子说明你想要什么吗?当从 B 调用活动 C 并获得例如数字“6”时,您希望从任何未来的随机计数中获得“6”,对吗?
  • 没错,当我从 B 反复去 C 时,我想要 6 个。

标签: android random shuffle


【解决方案1】:

在活动B的OnCreate中创建随机数的集合。发送,这样集合就不会重新启动了。发送列表中的号码

int random = randomNumber.get(i++);

【讨论】:

    猜你喜欢
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 2012-07-19
    • 2010-11-16
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    相关资源
    最近更新 更多