java 集合操作api

java.util.List:

1.List.addAll()：添加所有元素到列表中

com.sun.tools.javac.util.List:

1. 将指定的元素转为集合 

List strings = List.of("a", "bb", "cc", "ddd");

2. 比较两个集合所含元素是否相同

List strings = List.of("a", "bb", "cc", "ddd");
List strings2 = List.of("a", "bb", "cc","ddd");
boolean equals = List.equals(strings, strings2);
System.out.println(equals); //true

3.过滤掉集合里某个元素

List list1 = List.filter(List.of("a", "bb", "cc", "ddd"), "a");

System.out.println(list1.toString()); // bb,cc,ddd

4. 将list转为指定类型, 方法里面用的instance of进行判断

List strings = List.of("a", "bb", "cc", "ddd");
List convert = List.convert(String.class, strings);
System.out.println(convert);

 

java.util.Collections

1. Collections.unmodifiable*()  将list/map转换为不可变的

List strings = List.of("a", "bb", "cc", "ddd");
LinkedHashMap> collect1 = strings.stream().collect(groupingBy(String::length, LinkedHashMap::new, toCollection(TreeSet::new)));

Map> integerTreeSetMap = Collections.unmodifiableMap(collect1);
integerTreeSetMap.put(1,null); // 报java.lang.UnsupportedOperationException

2.Collections.enumeration 获取一个枚举集合

List list = new ArrayList(); 

list.add("aa"); 

Enumeration enumeration = Collections.enumeration(list);

java.util.stream

1. 并行流下,按顺序执行 forEachOrdered()

//输出的顺序不一定（效率更高）
Stream.of("AAA", "BBB", "CCC").parallel().forEach(s -> System.out.println("Output:" + s));
//输出的顺序与元素的顺序严格一致
Stream.of("AAA", "BBB", "CCC").parallel().forEachOrdered(s -> System.out.println("Output:" + s));

2.转成数组 toArray()

Person[] men = people.stream()
                          .filter(p -> p.getGender() == MALE)
                          .toArray(Person[]::new);

3. collect 收集

List strings = List.of("a", "bb", "cc", "ddd");

ArrayList