如何在没有迭代器的情况下遍历Set / HashSet？

如何遍历一个Set/ HashSet不包含以下内容？

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

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);
}

或使用Java 8：

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

至少有六种其他方式可以迭代一个集合。以下是我所知道的：

方法一

// 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我用于示例的内容：

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

将您的集合转换成数组 也可以帮助您遍历元素：

Object[] array = set.toArray();

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

为了演示，请考虑以下集合，其中包含不同的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));

人物模型类

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语句具有一种为通过Collections和数组进行迭代而设计的形式，该形式有时被称为增强的for语句，可用于使循环更紧凑和更易于阅读。

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

2. Java 8-java.lang.Iterable.forEach（消费者）

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

您可以使用函数操作获得更简洁的代码

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

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

以下是一些有关如何迭代Set及其演奏的提示：

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);
    }
}

该代码不言自明。

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花费时间最长而花费Iterator最快。

枚举（？）：

Enumeration e = new Vector(set).elements();
while (e.hasMoreElements())
    {
        System.out.println(e.nextElement());
    }

另一种方式（java.util.Collections.enumeration（））：

for (Enumeration e1 = Collections.enumeration(set); e1.hasMoreElements();)
    {
        System.out.println(e1.nextElement());
    }

Java 8：

set.forEach(element -> System.out.println(element));

要么

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

但是，已经有很好的答案。这是我的答案：

1. set.stream().forEach(System.out::println); // It simply uses stream to display set values
2. set.forEach(System.out::println); // It uses Enhanced forEach to display set values

另外，如果此Set为Custom类类型，例如：Customer。

Set<Customer> setCust = new HashSet<>();
    Customer c1 = new Customer(1, "Hena", 20);
    Customer c2 = new Customer(2, "Meena", 24);
    Customer c3 = new Customer(3, "Rahul", 30);

setCust.add(c1);
setCust.add(c2);
setCust.add(c3);
    setCust.forEach((k) -> System.out.println(k.getId()+" "+k.getName()+" "+k.getAge()));

//客户类别：

class Customer{
private int id;
private String name;
private int age;

public Customer(int id,String name,int age){
this.id=id;
this.name=name;
this.age=age;
} // Getter, Setter methods are present.}