blocks|key|1369222|text|排序集：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369223|return+new+TreeSet(setIWantSorted);|code-block|syntax|javascript|1369224|或者：|1369225|return+new+ArrayList(new+TreeSet(setIWantSorted));|1369226|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Sorted set: 

<pre><code>return new TreeSet(setIWantSorted);
</code></pre>

or:

<pre><code>return new ArrayList(new TreeSet(setIWantSorted));
</code></pre>

blocks|key|1369684|text|下面是如何使用Java+8的Streams完成此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369685|mySet.stream().sorted().collect(Collectors.toList());|code-block|syntax|javascript|1369686|或使用自定义比较器：|1369687|mySet.stream().sorted(myComparator).collect(Collectors.toList());|1369688|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here's how you can do it with Java 8's Streams:

<pre><code>mySet.stream().sorted().collect(Collectors.toList());
</code></pre>

or with a custom comparator:

<pre><code>mySet.stream().sorted(myComparator).collect(Collectors.toList());
</code></pre>

blocks|key|1584421|text|List+myList+=+new+ArrayList(collection);
Collections.sort(myList);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1584422|…然而，应该可以做到这一点。在适用的情况下添加泛型的风格。|unstyled|1584423|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>List myList = new ArrayList(collection);
Collections.sort(myList);
</code></pre>

… should do the trick however. Add flavour with Generics where applicable.

blocks|key|1369375|text|始终安全地使用比较器或可比较接口来提供排序实现(如果对象不是基本数据类型的字符串或包装类)。作为根据姓名对雇员进行排序的比较器实现的示例|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369376|++++List<Employees>+empList+=+new+LinkedList<Employees>(EmpSet);

++++class+EmployeeComparator+implements+Comparator<Employee>+{

++++++++++++public+int+compare(Employee+e1,+Employee+e2)+{
++++++++++++++++return+e1.getName().compareTo(e2.getName());
++++++++++++}

++++++++}

+++Collections.sort(empList+,+new+EmployeeComparator+());|code-block|syntax|javascript|1369377|当你需要在同一个对象上有不同的排序算法时，比较器是很有用的(比如emp名称，emp薪水等)。单模式排序可以通过在所需对象中使用可比较接口来实现。|1369378|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Always safe to use either Comparator or Comparable interface to provide sorting implementation (if the object is not a String or Wrapper classes for primitive data types) . 
As an example for a comparator implementation to sort employees based on name

<pre><code> List&lt;Employees&gt; empList = new LinkedList&lt;Employees&gt;(EmpSet);

 class EmployeeComparator implements Comparator&lt;Employee&gt; {

 public int compare(Employee e1, Employee e2) {
 return e1.getName().compareTo(e2.getName());
 }

 }

 Collections.sort(empList , new EmployeeComparator ());
</code></pre>

Comparator is useful when you need to have different sorting algorithm on same object (Say emp name, emp salary, etc). Single mode sorting can be implemented by using Comparable interface in to the required object.

blocks|key|1369193|text|没有单一的方法可以做到这一点。使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369194|@SuppressWarnings("unchecked")
public+static+<T+extends+Comparable>+List<T>+asSortedList(Collection<T>+collection)+{
++T[]+array+=+collection.toArray(
++++(T[])new+Comparable[collection.size()]);
++Arrays.sort(array);
++return+Arrays.asList(array);
}|code-block|syntax|javascript|1369195|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

There's no single method to do that. Use this:

<pre><code>@SuppressWarnings("unchecked")
public static &lt;T extends Comparable&gt; List&lt;T&gt; asSortedList(Collection&lt;T&gt; collection) {
 T[] array = collection.toArray(
 (T[])new Comparable[collection.size()]);
 Arrays.sort(array);
 return Arrays.asList(array);
}
</code></pre>

blocks|key|1584467|text|您可以将集合转换为ArrayList，其中可以使用Collections.sort(List)对ArrayList进行排序。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1584468|代码如下：|1584469|keySet+=+(Set)+map.keySet();
ArrayList+list+=+new+ArrayList(keySet);+++++
Collections.sort(list);|code-block|syntax|javascript|1584470|entityMap^0|9|9|P|M|1C|9|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

You can convert a set into an <code>ArrayList</code>, where you can sort the <code>ArrayList</code> using <code>Collections.sort(List)</code>.

Here is the code:

<pre><code>keySet = (Set) map.keySet();
ArrayList list = new ArrayList(keySet); 
Collections.sort(list);
</code></pre>

blocks|key|1369610|text|TreeSet+sortedset+=+new+TreeSet();
sortedset.addAll(originalset);

list.addAll(sortedset);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1369611|其中originalset+=未排序的集合，list+=要返回的列表|unstyled|1369612|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>TreeSet sortedset = new TreeSet();
sortedset.addAll(originalset);

list.addAll(sortedset);
</code></pre>

where originalset = unsorted set and list = the list to be returned

blocks|key|1584827|text|@Jeremy+Stein我想实现同样的代码。我还想对set+to+List进行排序，所以我没有使用Set，而是将set的值转换为list，并根据它的一个变量对列表进行排序。这段代码对我很有帮助，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584828|set.stream().sorted(Comparator.comparing(ModelClassName::sortingVariableName)).collect(Collectors.toList());|code-block|syntax|javascript|1584829|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

@Jeremy Stein I wanted to implement same code. As well I wanted to sort the set to list, So instead of using Set I converted set values into List and sort that list by it's one the variable.
This code helped me,

<pre><code>set.stream().sorted(Comparator.comparing(ModelClassName::sortingVariableName)).collect(Collectors.toList());
</code></pre>

blocks|key|1584873|text|我正在使用这段代码，我发现它比上面公认的答案更实用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584874|List<Thing>+thingList+=+new+ArrayList<>(thingSet);
thingList.sort((thing1,+thing2)+->+thing1.getName().compareToIgnoreCase(thing2.getName()));|code-block|syntax|javascript|1584875|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I am using this code, which I find more practical than the accepted answer above:

<pre><code>List&lt;Thing&gt; thingList = new ArrayList&lt;&gt;(thingSet);
thingList.sort((thing1, thing2) -&gt; thing1.getName().compareToIgnoreCase(thing2.getName()));
</code></pre>

In Java, I have a <code>Set</code>, and I want to turn it into a sorted <code>List</code>. Is there a method in the <code>java.util.Collections</code> package that will do this for me?

How do I sort a Set to a List in Java?

Java

在Java语言中，我有一个Set，我想把它变成一个排序的List。在java.util.Collections包中有没有方法可以帮我做到这一点呢？

问在Java中如何将集合排序到列表中？
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