我认为以上都不能回答这个问题，原因如下:

Since same functionality can be achieved by using other collections such as TreeSet, Collections, PriorityQueue..etc (but this is an alternative which will also impose their constraints i.e. Set will remove duplicate elements. Simply saying even if it does not impose any constraint, it does not answer the question why SortedList was not created by java community) Since List elements do not implements compare/equals methods (This holds true for Set & Map also where in general items do not implement Comparable interface but when we need these items to be in sorted order & want to use TreeSet/TreeMap,items should implement Comparable interface) Since List uses indexing & due to sorting it won't work (This can be easily handled introducing intermediate interface/abstract class)

但没有人告诉它背后的确切原因，因为我相信这类问题最好由java社区自己来回答，因为它只有一个具体的答案，但让我尽我所能回答如下:

我们知道排序是一个昂贵的操作，列表和Set/Map之间有一个基本的区别，列表可以有重复，而Set/Map不能。 这就是我们为Set/Map提供TreeSet/TreeMap形式的默认实现的核心原因。在内部，这是一个红黑树，每个操作(插入/删除/搜索)的复杂度为O(log N)，其中由于重复列表无法适合这种数据存储结构。

Now the question arises we could also choose a default sorting method for List also like MergeSort which is used by Collections.sort(list) method with the complexity of O(N log N). Community did not do this deliberately since we do have multiple choices for sorting algorithms for non distinct elements like QuickSort, ShellSort, RadixSort...etc. In future there can be more. Also sometimes same sorting algorithm performs differently depending on the data to be sorted. Therefore they wanted to keep this option open and left this on us to choose. This was not the case with Set/Map since O(log N) is the best sorting complexity.

如果你正在寻找一种方法来排序元素，但也能够以一种有效的方式通过索引访问它们，你可以做以下事情:

使用随机访问列表进行存储(例如ArrayList) 确保它总是有序的

然后，要添加或删除元素，可以使用集合。binarySearch获取插入/删除索引。因为您的列表实现了随机访问，所以您可以使用确定的索引有效地修改列表。

例子:

/**
 * @deprecated
 *      Only for demonstration purposes. Implementation is incomplete and does not 
 *      handle invalid arguments.
 */
@Deprecated
public class SortingList<E extends Comparable<E>> {
    private ArrayList<E> delegate;
    
    public SortingList() {
        delegate = new ArrayList<>();
    }
    
    public void add(E e) {
        int insertionIndex = Collections.binarySearch(delegate, e);
        
        // < 0 if element is not in the list, see Collections.binarySearch
        if (insertionIndex < 0) {
            insertionIndex = -(insertionIndex + 1);
        }
        else {
            // Insertion index is index of existing element, to add new element 
            // behind it increase index
            insertionIndex++;
        }
        
        delegate.add(insertionIndex, e);
    }
    
    public void remove(E e) {
        int index = Collections.binarySearch(delegate, e);
        delegate.remove(index);
    }
    
    public E get(int index) {
        return delegate.get(index);
    }
}

(在这个答案中可以看到更完整的实现)

JavaFX 排序列表

虽然花了一些时间，但Java 8确实有一个排序的List。 http://docs.oracle.com/javase/8/javafx/api/javafx/collections/transformation/SortedList.html

正如您在javadocs中所看到的，它是JavaFX集合的一部分，旨在提供一个ObservableList上的排序视图。

更新:注意，在Java 11中，JavaFX工具包已经移到JDK之外，现在是一个独立的库。JavaFX 11可以作为可下载的SDK或从MavenCentral获得。参见https://openjfx.io

我们有Collections.sort(arr)方法，它可以帮助对ArrayList arr进行排序。要以desc方式排序，我们可以使用集合。排序(arr Collections.reverseOrder ())

https://github.com/geniot/indexed-tree-map

考虑使用索引树映射。它是一个增强的JDK的TreeSet，它提供了通过索引访问元素的功能，并且无需迭代或隐藏的底层列表来备份树，就可以找到元素的索引。该算法基于每次有变化时更新已更改节点的权重。

对于任何新手来说，从2015年4月开始，Android现在在支持库中有一个SortedList类，专门用于与RecyclerView一起工作。这是关于它的博客文章。