java - 获取 10 个不重复的随机整数？

Question

这个问题已经有答案了:

How can I prevent the overlapping random numbers (8 个回答)

已关闭 8 年前。

我正在开发一个小程序，该程序可以从一副 52 张卡片中随机显示 10 张卡片。我使用 for 循环循环浏览 10 张卡片，并从我下载的图像中随机给他们一张卡片。当我编译并运行小程序时，有时会出现重复。有人可以给我任何指示，告诉我如何在不重复的情况下解决这个问题吗？

import java.applet.Applet;
import java.awt.Graphics;
import java.awt.Image;
import java.awt.Color; 
import java.util.Random;

public class Unit12Assignment1 extends Applet
{ 
     Image card[] = new Image[10];

public void init()
{
    String deckImages[] = {"cards_gif/c1.gif", "cards_gif/c2.gif", "cards_gif/c3.gif", "cards_gif/c4.gif",
        "cards_gif/c5.gif", "cards_gif/c6.gif", "cards_gif/c7.gif", "cards_gif/c8.gif", "cards_gif/c9.gif", 
        "cards_gif/c10.gif", "cards_gif/cj.gif", "cards_gif/ck.gif", "cards_gif/cq.gif",
        "cards_gif/s1.gif", "cards_gif/s2.gif", "cards_gif/s3.gif", "cards_gif/s4.gif",
        "cards_gif/s5.gif", "cards_gif/s6.gif", "cards_gif/s7.gif", "cards_gif/s8.gif",
        "cards_gif/s9.gif", "cards_gif/s10.gif", "cards_gif/sj.gif", "cards_gif/sk.gif",
        "cards_gif/sq.gif", "cards_gif/d1.gif", "cards_gif/d2.gif", "cards_gif/d3.gif",
        "cards_gif/d4.gif", "cards_gif/d5.gif", "cards_gif/d6.gif", "cards_gif/d7.gif",
        "cards_gif/d8.gif", "cards_gif/d9.gif", "cards_gif/d10.gif", "cards_gif/dj.gif",
        "cards_gif/dk.gif", "cards_gif/dq.gif", "cards_gif/h1.gif", "cards_gif/h2.gif",
        "cards_gif/h3.gif", "cards_gif/h4.gif", "cards_gif/h5.gif", "cards_gif/h6.gif",
        "cards_gif/h7.gif", "cards_gif/h8.gif", "cards_gif/h9.gif", "cards_gif/h10.gif",
        "cards_gif/hj.gif", "cards_gif/hk.gif", "cards_gif/hq.gif"};

    for( int i = 0; i < card.length; i++ )
    {
        int x = (int)(Math.random() * 52);
        card[i] = getImage(getDocumentBase(), deckImages[x]);
    }

}

public void paint(Graphics g)
{
    setBackground( Color.green );
    int x = -60;

    for( int i = 0; i < card.length / 2; i++ )
    {
        x = x + 90;
        g.drawImage(card[i], x, 30, this);
    }

    x = -60;

    for( int i = card.length / 2; i < card.length; i++ )
    {
        x = x + 90;
        g.drawImage(card[i], x, 150, this);
    }
}
}

Answer 1

有多种可能性。您可以检测重复性，并在这种情况下生成新索引，直到获得唯一的索引。其他选项可以是打乱索引数组，然后取前 10 个索引。发挥创意，这不是一项艰巨的任务。

另一种可能性:

// TODO initialize this to -1
int chosen[] = new int[10];
    for( int i = 0; i < card.length; i++ )
    {
      int x = (int)(Math.random() * 52);

      for(int j=0; j< chosen.length; j++)
        if(x==chosen[j])
        {
          x++;
          j=0;
          if(x>51)
            x=0;
        }
        card[i] = getImage(getDocumentBase(), deckImages[x]);
        chosen[i] = x;
    }

这不是严格随机的，但即使您遇到实现如下的随机数生成器，您也可以确定它确实会结束:return 42;