排序LinkedHashMap与值和时间优先级相等的情况下

Question

我知道这个主题可以很好地覆盖，但他们只使用一个只使用一个值的树图，如果相同的值不返回我想要的。

这是我的对象:

public classe myObjectToMap
{
    long id;
    double price;
    String name;
    long date;
    public static void GetMyObject(long id, double price, String name)
    {
        myObjectToMap newObject = new myObjectToMap;
        newObject.id = id;
        newObject.price = price;
        newObject.name = name;
        newObject.date = new Date().getTime();
        PlaceOfMyMap.myMap.put(id, newObject);
    }

这是我的地图所在的地方，我放了一个LinkedHashMap，我不知道这是Hashmap, TreeMap和LinkedHashMap之间最好的，我确实看到TreeMap给出了值的比较器，但我没有到达与多个值进行比较。

public class PlaceOfMyMap
{
    public static LinkedHashmap<Long, myObjectToMap> myMap = new LinkedHashmap<~>;
}

最后，这是我的主程序:

public class MainClass
{
    public static void main(String args[]) throws Exception
    {
        MyObjectToHashmap.GetMyObject(1, 26, "Mat")
        MyObjectToHashmap.GetMyObject(4, 25, "Tommy")
        MyObjectToHashmap.GetMyObject(16, 24, "Kate")
        MyObjectToHashmap.GetMyObject(63, 26, "Mary")
        MyObjectToHashmap.GetMyObject(99, 24, "Ronny")
 }

}

首先:我想按照时间优先级从最高价格到最低价格对它们进行排序，这意味着我希望Mat是第一个，Kate是第二个。

第二:我想按照时间优先级从最低价格到最高价格对它们进行排序，这意味着我希望Kate是第一个，Ronny是第二个。

有什么正确排序的技巧吗?

Answer 1

试试下面的代码。您可以反转compare()方法以获得反向排序

import java.util.Collections;
import java.util.Comparator;
import java.util.Date;
import java.util.LinkedHashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
class MyObject
{
    private long id;
    private double price;
    private String name;
    private long date;
    public MyObject(long id, double price, String name) {
    this.id = id;
    this.price = price;
    this.name = name;
    this.date = new Date().getTime();
    }
    @Override
    public String toString() {
        return "MyObject [id=" + id + ", price=" + price + ", name=" + name
                + "]";
    }
    public double getPrice() {
        return price;
    }
    public long getDate() {
        return date;
    }
}
public class MapSort {
    public static void main(String[] args) {
        Map<Long, MyObject> myMap = new LinkedHashMap<Long, MyObject>();
        myMap.put(1L, new MyObject(1, 26, "Mat"));
        myMap.put(4L, new MyObject(4, 25, "Tommy"));
        myMap.put(16L, new MyObject(16, 24, "Kate"));
        myMap.put(63L, new MyObject(63, 26, "Mary"));
        myMap.put(99L, new MyObject(99, 24, "Ronny"));
        System.out.println("Before Sorting");
        System.out.println(myMap);
        System.out.println("nAfter Sorting");
        System.out.println(sortMap(myMap));
    }
    private static Map<Long, MyObject> sortMap(
            Map<Long, MyObject> unsortedMap) {
        List<Entry<Long, MyObject>> list = new LinkedList<Entry<Long, MyObject>>(unsortedMap.entrySet());
        Collections.sort(list,
                new Comparator<Entry<Long, MyObject>>() {
                    @Override
                    public int compare(Entry<Long, MyObject> o1, Entry<Long, MyObject> o2) {
                        int priceResult = Double.valueOf(o1.getValue().getPrice()).compareTo(Double.valueOf(o2.getValue().getPrice()));
                        if(priceResult != 0) return priceResult;
                        return Long.valueOf(o1.getValue().getDate()).compareTo(Long.valueOf(o2.getValue().getDate()));
                    }
                });
        Map<Long, MyObject> sortedMap = new LinkedHashMap<Long, MyObject>();
        for(Entry<Long, MyObject> item : list){
            sortedMap.put(item.getKey(), item.getValue());
        }
        return sortedMap;
    }
}

Before Sorting
{1=MyObject [id=1, price=26.0, name=Mat], 4=MyObject [id=4, price=25.0, name=Tommy], 16=MyObject [id=16, price=24.0, name=Kate], 63=MyObject [id=63, price=26.0, name=Mary], 99=MyObject [id=99, price=24.0, name=Ronny]}
After Sorting
{16=MyObject [id=16, price=24.0, name=Kate], 99=MyObject [id=99, price=24.0, name=Ronny], 4=MyObject [id=4, price=25.0, name=Tommy], 1=MyObject [id=1, price=26.0, name=Mat], 63=MyObject [id=63, price=26.0, name=Mary]}