Java哈希集中元素的集合排序

Question

为什么第二组和第三组保持顺序：

Integer[] j = new Integer[]{3,4,5,6,7,8,9};
LinkedHashSet<Integer> i = new LinkedHashSet<Integer>();
Collections.addAll(i,j);
System.out.println(i); 

HashSet<Integer> hi = new HashSet<Integer>(i);
System.out.println(hi); 

LinkedHashSet<Integer> o = new LinkedHashSet<Integer>(hi);
System.out.println(o); 

以下是我得到的输出：

3,4,5,6,7,8,9
3,4,5,6,7,8,9
3,4,5,6,7,8,9

Answer 1

第二个（仅使用HashSet）只是巧合。从JavaDocs开始：

This class implements the Set interface, backed by a hash table (actually a HashMap instance). It makes no guarantees as to the iteration order of the set; in particular, it does not guarantee that the order will remain constant over time. This class permits the null element.

第三个（LinkedHashSet）是designed是这样的：

Hash table and linked list implementation of the Set interface, with predictable iteration order. This implementation differs from HashSet in that it maintains a doubly-linked list running through all of its entries. This linked list defines the iteration ordering, which is the order in which elements were inserted into the set (insertion-order). Note that insertion order is not affected if an element is re-inserted into the set. (An element e is reinserted into a set s if s.add(e) is invoked when s.contains(e) would return true immediately prior to the invocation.)

Answer 2

@Behrang的答案是好的，但更具体地说，HashSet似乎与LinkedHashSet顺序相同的唯一原因是integer.hashCode()恰好是整数值本身，所以在HashSet内部存储器中，数字恰好是顺序的。这是高度特定于实现的，正如@Behrang所说，这真是巧合

例如，如果使用new HashSet<>(4)将桶的初始数量设置为4（而不是16），则可能会得到以下输出：

HashSet<Integer> hi = new HashSet<Integer>(4);
...
[3, 4, 5, 6, 7, 8, 9]
[8, 9, 3, 4, 5, 6, 7]
[8, 9, 3, 4, 5, 6, 7]

如果你坚持价值观>；=16.你可能会得到这样的结果：

Integer[] j = new Integer[] { 3, 4, 5, 6, 7, 8, 9, 16 };
...
[3, 4, 5, 6, 7, 8, 9, 16]
[16, 3, 4, 5, 6, 7, 8, 9]
[16, 3, 4, 5, 6, 7, 8, 9]