如何在没有迭代器的情况下迭代Set/HashSet？

Question

如何迭代Set/ HashSet没有以下内容？

Iterator iter = set.iterator();
while (iter.hasNext()) {
    System.out.println(iter.next());
}

Answer 1

您可以使用增强的for循环:

Set<String> set = new HashSet<String>();

//populate set

for (String s : set) {
    System.out.println(s);
}

或者使用Java 8:

set.forEach(System.out::println);

Answer 2

至少有六种方法可以迭代一组.以下是我所知道的:

方法1

// Obsolete Collection
Enumeration e = new Vector(movies).elements();
while (e.hasMoreElements()) {
  System.out.println(e.nextElement());
}

方法2

for (String movie : movies) {
  System.out.println(movie);
}

方法3

String[] movieArray = movies.toArray(new String[movies.size()]);
for (int i = 0; i < movieArray.length; i++) {
  System.out.println(movieArray[i]);
}

方法4

// Supported in Java 8 and above
movies.stream().forEach((movie) -> {
  System.out.println(movie);
});

方法5

// Supported in Java 8 and above
movies.stream().forEach(movie -> System.out.println(movie));

方法6

// Supported in Java 8 and above
movies.stream().forEach(System.out::println);

这是HashSet我用于我的例子:

Set<String> movies = new HashSet<>();
movies.add("Avatar");
movies.add("The Lord of the Rings");
movies.add("Titanic");

Answer 3

将您的集转换为数组 也可以帮助您迭代元素:

Object[] array = set.toArray();

for(int i=0; i<array.length; i++)
   Object o = array[i];

Answer 4

要演示,请考虑以下集合,它包含不同的Person对象:

Set<Person> people = new HashSet<Person>();
people.add(new Person("Tharindu", 10));
people.add(new Person("Martin", 20));
people.add(new Person("Fowler", 30));

人模型类

public class Person {
    private String name;
    private int age;

    public Person(String name, int age) {
        this.name = name;
        this.age = age;
    }

    //TODO - getters,setters ,overridden toString & compareTo methods

}

1. for语句有一个用于迭代集合和数组的表单.这种形式有时被称为增强for语句,可用于使循环更紧凑和易于阅读.

for(Person p:people){
  System.out.println(p.getName());
}

2. Java 8 - java.lang.Iterable.forEach(Consumer)

people.forEach(p -> System.out.println(p.getName()));

default void forEach(Consumer<? super T> action)

Performs the given action for each element of the Iterable until all elements have been processed or the action throws an exception. Unless otherwise specified by the implementing class, actions are performed in the order of iteration (if an iteration order is specified). Exceptions thrown by the action are relayed to the caller. Implementation Requirements:

The default implementation behaves as if: 

for (T t : this)
     action.accept(t);

Parameters: action - The action to be performed for each element

Throws: NullPointerException - if the specified action is null

Since: 1.8

Answer 5

以下是有关如何迭代Set及其性能的一些提示:

public class IterateSet {

    public static void main(String[] args) {

        //example Set
        Set<String> set = new HashSet<>();

        set.add("Jack");
        set.add("John");
        set.add("Joe");
        set.add("Josh");

        long startTime = System.nanoTime();
        long endTime = System.nanoTime();

        //using iterator
        System.out.println("Using Iterator");
        startTime = System.nanoTime();
        Iterator<String> setIterator = set.iterator();
        while(setIterator.hasNext()){
            System.out.println(setIterator.next());
        }
        endTime = System.nanoTime();
        long durationIterator = (endTime - startTime);


        //using lambda
        System.out.println("Using Lambda");
        startTime = System.nanoTime();
        set.forEach((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationLambda = (endTime - startTime);


        //using Stream API
        System.out.println("Using Stream API");
        startTime = System.nanoTime();
        set.stream().forEach((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationStreamAPI = (endTime - startTime);


        //using Split Iterator (not recommended)
        System.out.println("Using Split Iterator");
        startTime = System.nanoTime();
        Spliterator<String> splitIterator = set.spliterator();
        splitIterator.forEachRemaining((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationSplitIterator = (endTime - startTime);


        //time calculations
        System.out.println("Iterator Duration:" + durationIterator);
        System.out.println("Lamda Duration:" + durationLambda);
        System.out.println("Stream API:" + durationStreamAPI);
        System.out.println("Split Iterator:"+ durationSplitIterator);
    }
}

代码是自我解释的.

持续时间的结果是:

Iterator Duration: 495287
Lambda Duration: 50207470
Stream Api:       2427392
Split Iterator:    567294

我们可以看到Lambda最长的时间Iterator是最快的.

Answer 6

您可以使用功能操作来获得更整洁的代码

Set<String> set = new HashSet<String>();

set.forEach((s) -> {
     System.out.println(s);
});