HashSet是Set接口的实现，元素无序、不可重复，==底层是一个HashMap==，用以保存数据。
不能保证元素的排列顺序，顺序有可能发生变化。
线程不安全。
集合元素可以是null，但只存在一个null。
线程安全： ==HashSet是线程不安全的==，需要用 Collections.synchronizedSet() 对其进行包装。

案例一

 @Test
    public void HashSetTest1() {
        HashSet<String> set = new HashSet<>();
        set.add("照子龙");
        set.add("官域");
        set.add("草草");
        set.add("流被");
        set.add("流被2");
        Iterator<String> iterator = set.iterator();
        while (iterator.hasNext()){
            String next = iterator.next();
            if(next=="流被"){
                iterator.remove();
            }
            System.out.println("结果："+next);
        }
        System.out.println("大小："+set.size());
    }

案例三线程不安全

  @Test
    public void HashSetTest2() throws InterruptedException {
        HashSet<String> set = new HashSet<>();
        //线程T-1
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 1; i <=500; i++) {
               set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-1").start();
        //线程T-2
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 495; i <= 1000; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-2").start();

       
        Thread.sleep(1000);
      
        System.out.println("大小：" + set.size());
    }

案例三：线程安全

 @Test
    public void HashSetTest2() throws InterruptedException {
       // HashSet<String> set = new HashSet<>();
        Set<String> set  = Collections.synchronizedSet(new HashSet<>());
        //线程T-1
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 1; i <=500; i++) {
               set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-1").start();
        //线程T-2
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 495; i <= 1000; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-2").start();

       
        Thread.sleep(1000);

        System.out.println("大小：" + set.size());
    }

二、LinkedHashSet

是Set接口的实现，==继承于HashSet==，元素不可重复，==底层是一个LinkedMap==，用以保存数据。
**==可保证元素的插入顺序。==**

是线程不安全的，需要用 Collections.synchronizedSet() 对其进行包装。

与HashSet的区别

LinkedHashSet在迭代访问Set中的全部元素时，性能比HashSet好，但是插入时性能稍微逊色于HashSet。即**==顺序访问性能好，插入、删除性能差。==**

案例一

  @Test
    public void LinkedHashSet() {
        LinkedHashSet<String> set = new LinkedHashSet<>();
        set.add("照子龙");
        set.add("官域");
        set.add("草草");
        set.add("流被");
        set.add("流被2");
        Iterator<String> iterator = set.iterator();
        while (iterator.hasNext()){
            String next = iterator.next();
            if(next=="流被"){
                iterator.remove();
            }
            System.out.println("结果："+next);
        }
        System.out.println("大小："+set.size());
    }

案例二

   @Test
    public void LinkSetTest1() throws InterruptedException {
        LinkedHashSet<String> set = new LinkedHashSet<>();
      //线程T-1
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 1; i <=500; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-1").start();
        //线程T-2
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 495; i <= 1000; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-2").start();

        Thread.sleep(1000);

        System.out.println("大小：" + set.size());
    }

案例三线程安全

 @Test
    public void LinkSetTest1() throws InterruptedException {
        //LinkedHashSet<String> set = new LinkedHashSet<>();
        Set<String> set  = Collections.synchronizedSet(new LinkedHashSet<>());

        //线程T-1
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 1; i <=500; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-1").start();
        //线程T-2
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 495; i <= 1000; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-2").start();

        Thread.sleep(1000);

        System.out.println("大小：" + set.size());
    }

三、TreeSet

TreeSet是SortedSet接口的唯一实现类，元素不可重复。==底层是一个TreeMap==，用以保存数据。
TreeSet**==默认可保证元素的自然顺序（元素需要实现Comparable接口），可指定排序规则（需要重写元素的hashCode()方法和equals()方法）==**。

==TreeSet是线程不安全的==，需要用 Collections.synchronizedSet() 对其进行包装。

TreeSet的性能比HashSet差但是我们在需要排序的时候可以用TreeSet因为他是自然排序也就是升序下面是TreeSet实现代码这个类也似只能通过迭代器迭代元素

案例一

  @Test
    public void TreeSet() {
        TreeSet<String> set = new TreeSet<>();
        set.add("照子龙");
        set.add("官域");
        set.add("草草");
        set.add("流被");
        set.add("流被2");
        Iterator<String> iterator = set.iterator();
        while (iterator.hasNext()){
            String next = iterator.next();
            if(next=="流被"){
                iterator.remove();
            }
            System.out.println("结果："+next);
        }
        System.out.println("大小："+set.size());
    }

案例二

 @Test
    public void TreeSet2() {
        TreeSet<String> set = new TreeSet<>();
        set.add("G");
        set.add("B");
        set.add("C");
        set.add("A");
        set.add("D");
        Iterator<String> iterator = set.iterator();
        while (iterator.hasNext()){
            String next = iterator.next();
          
            System.out.println("结果："+next);
        }
        System.out.println("大小："+set.size());
    }

案例三

  @Test
    public void TreeSet1() throws InterruptedException {
        TreeSet<String> set = new TreeSet<>();
        //线程T-1
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 1; i < 500; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-1").start();
        //线程T-2
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 498; i <= 1000; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-2").start();

        //线程T-3
       /* new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            Iterator<String> iterator = set.iterator();
            while (iterator.hasNext()){
                String next = iterator.next();
                if("a5".equals(next)){
                    iterator.remove();
                }
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-3").start();*/
        Thread.sleep(1000);

        System.out.println("大小：" + set.size());
    }

案例四线程安全

 @Test
    public void TreeSet1() throws InterruptedException {
       // TreeSet<String> set = new TreeSet<>();
       // ConcurrentSkipListSet<String> set = new ConcurrentSkipListSet<>();
        Set<String> set = Collections.synchronizedSet(new TreeSet<>());
        //线程T-1
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 1; i < 500; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-1").start();
        //线程T-2
        new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            for (int i = 498; i <= 1000; i++) {
                set.add("a"+i);
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-2").start();

        //线程T-3
       /* new Thread(() -> {
            System.out.println("当前线程：" + Thread.currentThread().getName());

            Iterator<String> iterator = set.iterator();
            while (iterator.hasNext()){
                String next = iterator.next();
                if("a5".equals(next)){
                    iterator.remove();
                }
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

        }, "T-3").start();*/
        Thread.sleep(1000);

        System.out.println("大小：" + set.size());
    }

总结：

posted on 2019-08-29 22:35 源无极阅读(120) 评论(0) 编辑收藏举报

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