Kotlin随机图像在其他图像视图中无重复

发布于 2025-02-02 04:08:58 字数 738 浏览 5 评论 0原文

我该如何解决Kotlin，我即将完成一个应用程序，但我无法关闭所有内容？

我不擅长的功能

是，使用三个ipaimViews，我希望ImageView中的映像不要在ImageView 2和3中重复相同的数字，因此我希望列表成为一个，并且没有重复的内容可以显示这三个图像视图之间是代码：

 private lateinit var listId: List<Int>
 
 initializeList()
 
 pickNumber ()

    }

    private fun pickNumber (){
        binding.imageView.setImageResource(listId.random())
        binding.imageView2.setImageResource(listId.random())
        binding.imageView3.setImageResource(listId.random())
        
        
         private fun initializeList() {
        listId = listOf(
            R.drawable.we ,
            R.drawable.wq,
            R.drawable.set,
            R.drawable.opt,
            R.drawable.ups,
        )

带有80张图像

原文

how can I solve in kotlin, I am about to finish an app but I can't close everything?

the function that I am not good at

the fact is that with three imageviews I would like the image in imageview not to repeat the same figure in imageview 2 and 3 so I would like the list to be one and that there are no duplicates to show between the three imageviews here is the code:

 private lateinit var listId: List<Int>
 
 initializeList()
 
 pickNumber ()

    }

    private fun pickNumber (){
        binding.imageView.setImageResource(listId.random())
        binding.imageView2.setImageResource(listId.random())
        binding.imageView3.setImageResource(listId.random())
        
        
         private fun initializeList() {
        listId = listOf(
            R.drawable.we ,
            R.drawable.wq,
            R.drawable.set,
            R.drawable.opt,
            R.drawable.ups,
        )

with 80 images

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甜扑 2025-02-09 04:08:58

正如Gpunto建议的那样，您可以先洗个清单，然后逐一获取元素。

private fun pickNumber (){
    val (id1, id2, id3) = list.shuffled()  
    binding.imageView.setImageResource(id1)
    binding.imageView2.setImageResource(id2)
    binding.imageView3.setImageResource(id3)
}

在这里，我使用列表破坏以获取列表的第三个元素。

官方文档

As gpunto suggested, you can shuffle the list first and then fetch elements one by one.

private fun pickNumber (){
    val (id1, id2, id3) = list.shuffled()  
    binding.imageView.setImageResource(id1)
    binding.imageView2.setImageResource(id2)
    binding.imageView3.setImageResource(id3)
}

Here I used list destructuring to get first three elements of the list.

Official docs

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