通用方法来查找3个值的中位数 [英] Generic Method to find the median of 3 values

问题描述

我需要一种方法来获得3个值的中位数，我认为这是一个很好的机会来编写一个通用的方法，因为我实际上没有这样做。我写了这篇文章，看起来非常直截了当，虽然我得到了一个警告，但根据我的测试，它似乎工作正常。

我知道我可以使用固有排序的集合，或 Collections.sort（），但这种方法是为了理解。

我想查明几件事：

1. 我注意到这不起作用，如果我试图声明 medianHelper 与 Arrays.asList（a，b，c）为什么会这样？试图搜索这个结果给了我无关的结果，因为我不确定发生了什么，所以它是难以捉摸的。我得到了一个 UnsupportedOperationException ，但这不是我下面的方式。
   
2. 为什么我会收到警告？什么是错误/丢失？
   

该方法如下：

  private static< T extends Comparable> T中位数（T a，T b，T c）{
 List< T> medianHelper = new ArrayList<>（）; 
 T max; 
 T min; 
 
 medianHelper.add（a）; 
 medianHelper.add（b）; 
 medianHelper.add（c）; 
 
 if（a.compareTo（b）> = 0）{
 max = a; 
 min = b; 
} else {
 max = b; 
 min = a; 
} 
 
 if（max.compareTo（c）== -1）{
 max = c; 
} 
 
 if（min.compareTo（c）> = 0）{
 min = c; 
} 
 
 medianHelper.remove（max）; 
 medianHelper.remove（min）; 
 
返回medianHelper.get（0）; 
 
  

解决方案

类型参数 T ，比如 Comparable 也是通用的。

它应该是：

  private static< T extends Comparable <？ super T>> T中位数（T a，T b，T c）
  

此外，您可以排序> medianHelper 列表，因为它的元素将> Comparable 。因此，您的方法可以显着缩短为：

  private static< T extends Comparable< ;? super T>> T中位数（T a，T b，T c）{
 List< T> medianHelper = Arrays.asList（a，b，c）; 
 
 Collections.sort（medianHelper）; 
 
 return medianHelper.get（1）; 
 
 $ / code> 

请注意 Arrays.asList（）返回一个不可修改的列表，这意味着您不允许在创建元素后添加/删除元素。如果你想自己进行比较，你可以使用新的ArrayList<> c $ c>来代替 Arrays.asList（）然后手动将元素添加到它。

I needed a method to get the median of 3 values, I thought it a good opportunity to write a generic method since I don't really have that practiced. I wrote this and it seems pretty straight-forward, though I get a warning, but it seems to work fine, according to my tests.

I'm aware I could use an inherently sorted set, or Collections.sort(), but this approach is for the sake of understanding.

I want to pinpoint a few things:

1. I noticed this doesn't work if I tried to declare medianHelper with Arrays.asList(a, b, c) why is this? Trying to search this gives me unrelated results and it's otherwise elusive since I'm not sure what is happening. I get an UnsupportedOperationException, but this is not present the way I have it below.
2. Why am I getting a warning? What is wrong/missing?

The method follows:

private static <T extends Comparable> T median(T a, T b, T c) {
    List<T> medianHelper = new ArrayList<>();
    T max;
    T min;

    medianHelper.add(a);
    medianHelper.add(b);
    medianHelper.add(c);

    if (a.compareTo(b) >= 0) {
        max = a;
        min = b;
    } else {
        max = b;
        min = a;
    }

    if (max.compareTo(c) == -1) {
        max = c;
    }

    if (min.compareTo(c) >= 0) {
        min = c;
    }

    medianHelper.remove(max);
    medianHelper.remove(min);

    return medianHelper.get(0);
}

解决方案

You haven't correctly introduced the type-parameter T, as Comparable is generic, too.

It should rather be:

private static <T extends Comparable<? super T>> T median(T a, T b, T c) 

Furthermore, you can just sort the medianHelper list, since its elements will be Comparable. So your method can be significantly shortened to:

private static <T extends Comparable<? super T>> T median(T a, T b, T c) {
    List<T> medianHelper = Arrays.asList(a, b, c);

    Collections.sort(medianHelper);

    return medianHelper.get(1);
}

Note that Arrays.asList() returns an unmodifiable list, which means you're not allowed to add/remove elements after it's created. If you wish to do the comparisons yourself, you can use new ArrayList<> instead of Arrays.asList() and then manually add the elements to it.

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