Set集合重复元素的判断

Set集合比较特殊,他的方法基本上和Collection接口的方法一致,并且他没有根据索引得到对应值的get(int index)方法. 更重要的是Set集合不能放置重复的元素.她是怎么判断的呢?那么,我们首先要让对象具有比较性
一,存储的自定义对象实现接口 Comparable
复写public int compareTo()方法. 返回0表示二者相等. 如果返回大于0的整数(一般写1),则认为该对象大于比较的对象.如果返回小于0的整数(一般是-1),则认为该对象小于比较的对象.

下面我们定义一个类Person,如果对象的年龄相等,我们就认为这两个对象相等

public static void main(String[] args) {
        testTreeSet();
    }

    private static void testTreeSet() {
        TreeSet treeSet = new TreeSet();
        treeSet.add(new Person("Tom", 20));
        treeSet.add(new Person("Jim", 21));
        treeSet.add(new Person("Jim", 20));

        Iterator it = treeSet.iterator();
        while (it.hasNext()) {
            Person p = it.next();
            System.out.println(p.toString());
        }
        System.out.println("size = "+treeSet.size());
    }

    static class Person implements Comparable {
        private String name;
        private int age;

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

        public String getName() {
            return name;
        }

        public void setName(String name) {
            this.name = name;
        }

        public int getAge() {
            return age;
        }

        public void setAge(int age) {
            this.age = age;
        }

        @Override
        public int hashCode() {
            System.out.println("hashcode  " + 17 * age);
            return 17 * age;
        }

        @Override
        public boolean equals(Object obj) {
            Person p = (Person) obj;
            boolean b = (this.age == p.getAge());
            if (b) {
                System.out.println("一样      " + this.toString() + "   "
                        + p.toString());
            } else {
                System.out.println("不一样      " + this.toString() + "   "
                        + p.toString());
            }
            return b;
        }

        @Override
        public String toString() {
            return "Person [name=" + name + ", age=" + age + "]";
        }

        @Override
        public int compareTo(Person o) {
            if (this.age == o.getAge()) {
                return 0;
            }

            if (this.age > o.getAge()) {
                return 1;
            }

            return -1;
        }

    }

运行结果为:
Person [name=Tom, age=20]
Person [name=Jim, age=21]
size = 2

由此分析,在操作treeSet.add(new Person(“Jim”, 20));的时候认为他是一个相同的对象,那么就无法再放入了

二,存储的自定义对象实现接口 Comparable

public static void main(String[] args) {
        testTreeSet();
    }

    private static void testTreeSet() {
        TreeSet treeSet = new TreeSet(new Comparator() {
            @Override
            public int compare(Person o1, Person o2) {
                if (o1.getAge() == o2.getAge()) {
                    return 0;
                }

                if (o1.getAge() > o2.getAge()) {
                    return 1;
                }

                return -1;
            }
        });
        treeSet.add(new Person("Tom", 20));
        treeSet.add(new Person("Jim", 21));
        treeSet.add(new Person("Jim", 20));

        Iterator it = treeSet.iterator();
        while (it.hasNext()) {
            Person p = it.next();
            System.out.println(p.toString());
        }
        System.out.println("size = "+treeSet.size());
    }

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

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

        public String getName() {
            return name;
        }

        public void setName(String name) {
            this.name = name;
        }

        public int getAge() {
            return age;
        }

        public void setAge(int age) {
            this.age = age;
        }

        @Override
        public String toString() {
            return "Person [name=" + name + ", age=" + age + "]";
        }
    }

结果是:
Person [name=Tom, age=20]
Person [name=Jim, age=21]
size = 2
存入了两个结果.
treeSet.add(new Person(“Jim”, 20)); Jim的年龄和Tom相同被认为是一样的对象