我正在使用9图像视图我想要随机设置图像到imageview,当我点击刷新按钮,但我尝试这样它是为图像的随机分配工作,但它一次在两个(或)三个imageview中重复相同的图像.我的代码中的问题在哪里..

 final int[] imageViews = {
            R.id.imgview11, R.id.imgview12, R.id.imgview13, 
            R.id.imgview21, R.id.imgview22, R.id.imgview23, 
            R.id.imgview31, R.id.imgview32, R.id.imgview33 };

    final int[] images = {
            R.drawable.i1, R.drawable.i2, R.drawable.i3, 
            R.drawable.i4, R.drawable.i5, R.drawable.i6, 
            R.drawable.i7, R.drawable.i8, R.drawable.empty };

    final ImageButton shuffle = (ImageButton) findViewById(R.id.new_puzzle); 
    shuffle.setOnClickListener(new View.OnClickListener() {
        public void onClick(View view) {        
            Random generator = new Random();
            //int n = 9;
            //n = generator.nextInt(n);
            //Random random = new Random(System.currentTimeMillis());
            for(int v : imageViews) {
                ImageView iv = (ImageView)findViewById(v);
                iv.setImageResource(images[generator.nextInt(images.length - 1)]);
            }
        }
    });      

我不想重复,一个图像仅适用于一个imageview ..

解决方法:

使用blessenm的帖子,我写了一个你需要的类似代码.检查这是否对您有所帮助.

shuffle.setOnClickListener(new View.OnClickListener() {
            public void onClick(View view) { 

            Random rng = new Random(); 
            List<Integer> generated = new ArrayList<Integer>();
            for (int i = 0; i < 9; i++)
            {
              while(true)
              {
                 Integer next = rng.nextInt(9) ;
                 if (!generated.contains(next))
                 {
                    generated.add(next);
                    ImageView iv = (ImageView)findViewById(imageViews[i]);
                    iv.setImageResource(images[next]);
                    break;
                 }
               }
            }
            }
        });