如何在 Java 中将 Set 排序为 List？
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How do I sort a Set to a List in Java?
提问by Jeremy Stein
In Java, I have a Set, and I want to turn it into a sorted List. Is there a method in the java.util.Collectionspackage that will do this for me?
在 Java 中，我有一个Set, 我想把它变成一个排序的List. java.util.Collections包中是否有方法可以为我执行此操作？
采纳答案by erickson
The answer provided by the OPis not the best. It is inefficient, as it creates a new Listandan unnecessary new array. Also, it raises "unchecked" warnings because of the type safety issues around generic arrays.
OP提供的答案并不是最好的。它效率低下，因为它创建了一个新的List和不必要的新数组。此外，由于泛型数组的类型安全问题，它会引发“未经检查”的警告。
Instead, use something like this:
相反，使用这样的东西：
public static
<T extends Comparable<? super T>> List<T> asSortedList(Collection<T> c) {
List<T> list = new ArrayList<T>(c);
java.util.Collections.sort(list);
return list;
}
Here's a usage example:
这是一个使用示例：
Map<Integer, String> map = new HashMap<Integer, String>();
/* Add entries to the map. */
...
/* Now get a sorted list of the *values* in the map. */
Collection<String> unsorted = map.values();
List<String> sorted = Util.asSortedList(unsorted);
回答by Jeremy Stein
There's no single method to do that. Use this:
没有单一的方法可以做到这一点。用这个：
@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);
}
回答by Steve B.
Sorted set:
排序集：
return new TreeSet(setIWantSorted);
or:
或者：
return new ArrayList(new TreeSet(setIWantSorted));
回答by Esko
List myList = new ArrayList(collection);
Collections.sort(myList);
… should do the trick however. Add flavour with Generics where applicable.
......但是应该可以解决问题。在适用的情况下使用泛型添加风味。
回答by Amit
You can convert a set into an ArrayList, where you can sort the ArrayListusing Collections.sort(List).
您可以将集合转换为ArrayList，您可以在其中对ArrayListusing进行排序Collections.sort(List)。
Here is the code:
这是代码：
keySet = (Set) map.keySet();
ArrayList list = new ArrayList(keySet);
Collections.sort(list);
回答by Satheesh Cheveri
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
始终安全地使用 Comparator 或 Comparable 接口来提供排序实现（如果对象不是原始数据类型的 String 或 Wrapper 类）。作为比较器实现根据姓名对员工进行排序的示例
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 ());
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.
当您需要对同一对象使用不同的排序算法时，比较器很有用（比如 emp 名称、emp 薪水等）。可以通过使用 Comparable 接口到所需对象中来实现单模式排序。
回答by Sujith Mohan
TreeSet sortedset = new TreeSet();
sortedset.addAll(originalset);
list.addAll(sortedset);
where originalset = unsorted set and list = the list to be returned
其中 originalset = unsorted set 和 list = 要返回的列表
回答by nschum
Here's how you can do it with Java 8's Streams:
以下是使用 Java 8 的 Streams 实现的方法：
mySet.stream().sorted().collect(Collectors.toList());
or with a custom comparator:
或使用自定义比较器：
mySet.stream().sorted(myComparator).collect(Collectors.toList());
回答by Deepak Kumbhar
@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,
@Jeremy Stein 我想实现相同的代码。同样，我想将集合排序到列表中，因此我没有使用 Set 我将集合值转换为 List 并通过它的一个变量对该列表进行排序。这段代码帮助了我，
set.stream().sorted(Comparator.comparing(ModelClassName::sortingVariableName)).collect(Collectors.toList());
回答by nn4l
I am using this code, which I find more practical than the accepted answer above:
我正在使用此代码，我发现它比上面接受的答案更实用：
List<Thing> thingList = new ArrayList<>(thingSet);
thingList.sort((thing1, thing2) -> thing1.getName().compareToIgnoreCase(thing2.getName()));
