java TreeSet:比较和相等 [英] java TreeSet: comparing and equality

问题描述

我想用属性"sort_1"对对象列表进行排序.但是，当我想要删除时，我希望它使用属性"id".以下代码代表了问题.

I'd like to have list of object sorted with property 'sort_1'. But when I want to remove I'd like it to use property 'id'. The following code represents the problem.

package javaapplication1;

import java.util.TreeSet;

public class MyObj implements Comparable<MyObj> {
    public long sort_1;
    public long id;

    public MyObj(long sort, long id) {
        this.sort_1=sort;
        this.id=id;
    }

    @Override
    public int compareTo(MyObj other) {        
        int ret = Long.compare(sort_1, other.sort_1);               
        return ret;
    }    

    public String toString() {
        return id+":"+sort_1;
    }

    public static void main(String[] args) {
        TreeSet<MyObj> lst=new TreeSet<MyObj>();

                MyObj o1 = new MyObj(99,1);
        MyObj o2 = new MyObj(11,9);

        lst.add(o1);
        lst.add(o2);    

        System.out.println(lst);

                MyObj o3 = new MyObj(1234, 1);
                //remove myObje with id 1
                boolean remove=lst.remove(o3);

                System.out.println(lst);
    }

}

此代码的输出是:

[9:11, 1:99]
[9:11, 1:99]

我需要对列表进行排序，因为我对列表做了很多添加.我不想明确使用任何排序"方法.我有什么选择?

I need to have list sorted as I do a lot of additions to the list. I don't want to explicitly use any 'sort' method. What are my options ?

我的要求是:具有"id"的对象是唯一的，但是可以存在具有重复"sort"值的对象.

My requirement is to have: objects with 'id' as unique but there can be object's with duplicate 'sort' value.

推荐答案

我认为您遇到的问题是您正在实现

I think the problem you're having is that you are implementing Comparable, but your implementation seems to be inconsistent with equals - and you have not implemented any equality methods. That is:

当且仅当e1.compareTo(e2)== 0具有与e1.equals(e2)相同的布尔值(对于C类的每个e1和e2)时，类C的自然顺序才与equals一致.

The natural ordering for a class C is said to be consistent with equals if and only if e1.compareTo(e2) == 0 has the same boolean value as e1.equals(e2) for every e1 and e2 of class C

在您的情况下，当您构建这三个对象时:

In your case, when you build these three objects:

MyObj o1 = new MyObj(99,1);
MyObj o2 = new MyObj(11,9);
MyObj o3 = new MyObj(1234, 1);

您会看到o1.compareTo(o3)== -1，而o1.equals(o3)== false.

You will see that o1.compareTo(o3) == -1, while o1.equals(o3) == false.

但是您似乎想要o1.equals(o3)== true.

But you seem to want o1.equals(o3) == true.

此外，请注意，如果对象已经存在于集合中，则TreeSet.add()返回false.此检查基于equals()方法.

Also, recognize that TreeSet.add() returns false if the object already exists in the set. This check is based on the equals() method.

要解决此问题，请重写Object.equals()和Object.hashCode()，使它们考虑MyObj.id字段，并在不相等时继续在compareTo()方法中使用sort_1字段.

To remedy this, override Object.equals() and Object.hashCode() such that they take into consideration the MyObj.id field, and continue to use the sort_1 field in the compareTo() method when they are not equal.

package javaapplication1;

import java.util.TreeSet;

public class MyObj implements Comparable<MyObj> {

    public long sort_1;
    public long id;

    public MyObj(long sort, long id) {
        this.sort_1 = sort;
        this.id = id;
    }

    @Override
    public int compareTo(MyObj other) {
        return (this.equals(other))? 0 : Long.compare(sort_1, other.sort_1);
    }

    @Override
    public boolean equals(Object obj) {
        MyObj other = (MyObj) obj;
        return this.id == other.id && this.sort_1 == other.sort_1;
    }

    @Override
    public int hashCode() {
        return (int) id;
    }


    public String toString() {
        return id + ":" + sort_1;
    }

    public static void main(String[] args) {
        TreeSet<MyObj> lst = new TreeSet<MyObj>();

        MyObj o1 = new MyObj(99L, 1L);
        MyObj o2 = new MyObj(11L, 9L);
        MyObj o3 = new MyObj(1234L, 1L);       
        MyObj o4 = new MyObj(1234L, 1L);   

        System.out.println( "Adding o1: " + lst.add(o1));
        System.out.println( "Adding o2: " + lst.add(o2));
        System.out.println( "Adding o3: " + lst.add(o3));
        System.out.println( "Adding o4: " + lst.add(o4));        

        System.out.println(lst);

        System.out.println("o1.compareTo(o3) : " + o1.compareTo(o3));
        System.out.println("o1.equals(o3) : " + o1.equals(o3));

        //remove myObje with id 1
        boolean remove = lst.remove(o3);

        System.out.println(lst);
    }

}

输出:

Adding o1: true
Adding o2: true
Adding o3: true
Adding o4: false
[9:11, 1:99, 1:1234]
o1.compareTo(o3) : -1
o1.equals(o3) : false
[9:11, 1:99]

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