我想创建一个用于测试的选项列表。起初,我这样做:
ArrayList<String> places = new ArrayList<String>();
places.add("Buenos Aires");
places.add("Córdoba");
places.add("La Plata");
然后,我将代码重构如下:
ArrayList<String> places = new ArrayList<String>(
Arrays.asList("Buenos Aires", "Córdoba", "La Plata"));
有更好的方法吗?
在Java 9中,我们可以很容易地在一行中初始化ArrayList:
List<String> places = List.of("Buenos Aires", "Córdoba", "La Plata");
or
List<String> places = new ArrayList<>(List.of("Buenos Aires", "Córdoba", "La Plata"));
Java 9的这种新方法与以前的方法相比有许多优点:
空间效率不可变性线程安全
有关更多详细信息,请参阅本文->List.of和Arrays.asList之间的区别是什么?
集合文本并没有进入Java 8,但可以使用流API在一行中初始化列表:
List<String> places = Stream.of("Buenos Aires", "Córdoba", "La Plata").collect(Collectors.toList());
如果您需要确保列表是ArrayList:
ArrayList<String> places = Stream.of("Buenos Aires", "Córdoba", "La Plata").collect(Collectors.toCollection(ArrayList::new));
使用Eclipse集合,您可以编写以下内容:
List<String> list = Lists.mutable.with("Buenos Aires", "Córdoba", "La Plata");
您还可以更具体地了解类型,以及它们是可变的还是不可变的。
MutableList<String> mList = Lists.mutable.with("Buenos Aires", "Córdoba", "La Plata");
ImmutableList<String> iList = Lists.immutable.with("Buenos Aires", "Córdoba", "La Plata");
您也可以对套装和包进行同样的操作:
Set<String> set = Sets.mutable.with("Buenos Aires", "Córdoba", "La Plata");
MutableSet<String> mSet = Sets.mutable.with("Buenos Aires", "Córdoba", "La Plata");
ImmutableSet<String> iSet = Sets.immutable.with("Buenos Aires", "Córdoba", "La Plata");
Bag<String> bag = Bags.mutable.with("Buenos Aires", "Córdoba", "La Plata");
MutableBag<String> mBag = Bags.mutable.with("Buenos Aires", "Córdoba", "La Plata");
ImmutableBag<String> iBag = Bags.immutable.with("Buenos Aires", "Córdoba", "La Plata");
注意:我是Eclipse集合的提交人。