我在java中有一个双精度的列表,我想按降序排序数组列表。

输入数组列表如下:

List<Double> testList = new ArrayList();

testList.add(0.5);
testList.add(0.2);
testList.add(0.9);
testList.add(0.1);
testList.add(0.1);
testList.add(0.1);
testList.add(0.54);
testList.add(0.71);
testList.add(0.71);
testList.add(0.71);
testList.add(0.92);
testList.add(0.12);
testList.add(0.65);
testList.add(0.34);
testList.add(0.62);

输出应该是这样的

0.92
0.9
0.71
0.71
0.71
0.65
0.62
0.54
0.5
0.34
0.2
0.12
0.1
0.1
0.1

当前回答

  yearList = arrayListOf()
    for (year in 1950 until 2021) {
        yearList.add(year)
    }

   yearList.reverse()
    val list: ArrayList<String> = arrayListOf()

    for (year in yearList) {
        list.add(year.toString())
    }

其他回答

集合。sort允许您传递一个定义排序逻辑的Comparator实例。因此,我们可以简单地传递Collections.reverseOrder()来排序,而不是按自然顺序对列表进行排序,然后反转它:

// import java.util.Collections;
Collections.sort(testList, Collections.reverseOrder());

正如@Marco13所提到的,除了更习惯(可能更有效),使用倒序比较器可以确保排序是稳定的(这意味着当元素的顺序根据比较器相等时,它们的顺序不会改变,而倒序将改变顺序)

//Here is sorted List alphabetically with syncronized

package com.mnas.technology.automation.utility;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.List;

import org.apache.log4j.Logger;

/**
 * @author manoj.kumar
 */
public class SynchronizedArrayList {
    static Logger log = Logger.getLogger(SynchronizedArrayList.class.getName());

    @SuppressWarnings("unchecked")
    public static void main(String[] args) {

        List<Employee> synchronizedList = Collections.synchronizedList(new ArrayList<Employee>());
        synchronizedList.add(new Employee("Aditya"));
        synchronizedList.add(new Employee("Siddharth"));
        synchronizedList.add(new Employee("Manoj"));
        Collections.sort(synchronizedList, new Comparator() {
            public int compare(Object synchronizedListOne, Object synchronizedListTwo) {
                //use instanceof to verify the references are indeed of the type in question
                return ((Employee) synchronizedListOne).name
                        .compareTo(((Employee) synchronizedListTwo).name);
            }
        }); 
    /*for( Employee sd : synchronizedList) {
    log.info("Sorted Synchronized Array List..."+sd.name);
    }*/

        // when iterating over a synchronized list, we need to synchronize access to the synchronized list
        synchronized (synchronizedList) {
            Iterator<Employee> iterator = synchronizedList.iterator();
            while (iterator.hasNext()) {
                log.info("Sorted Synchronized Array List Items: " + iterator.next().name);
            }
        }

    }
}

class Employee {
    String name;

    Employee(String name) {
        this.name = name;

    }
}

例如,我有一个类Person: String name, int age ==>构造函数new Person(name,age)

import java.util.Collections;
import java.util.ArrayList;
import java.util.Arrays;


public void main(String[] args){
    Person ibrahima=new Person("Timera",40);
    Person toto=new Person("Toto",35);
    Person alex=new Person("Alex",50);
    ArrayList<Person> myList=new ArrayList<Person>
    Collections.sort(myList, new Comparator<Person>() {
        @Override
        public int compare(Person p1, Person p2) {
            // return p1.age+"".compareTo(p2.age+""); //sort by age
            return p1.name.compareTo(p2.name); // if you want to short by name
        }
    });
    System.out.println(myList.toString());
    //[Person [name=Alex, age=50], Person [name=Timera, age=40], Person [name=Toto, age=35]]
    Collections.reverse(myList);
    System.out.println(myList.toString());
    //[Person [name=Toto, age=35], Person [name=Timera, age=40], Person [name=Alex, age=50]]

}

对于您的示例,这将在Java 8中发挥作用

List<Double> testList = new ArrayList();
testList.sort(Comparator.naturalOrder());

但是如果你想要排序对象的一些字段,你可以很容易地通过:

testList.sort(Comparator.comparing(ClassName::getFieldName));

or

 testList.sort(Comparator.comparing(ClassName::getFieldName).reversed());

or

 testList.stream().sorted(Comparator.comparing(ClassName::getFieldName).reversed()).collect(Collectors.toList());

来源:https://docs.oracle.com/javase/8/docs/api/java/util/Comparator.html

你可以这样做:

List<String> yourList = new ArrayList<String>();
Collections.sort(yourList, Collections.reverseOrder());

集合有一个默认的比较器可以帮助你。

另外,如果你想使用一些Java 8的新特性,你可以这样做:

List<String> yourList = new ArrayList<String>();
yourList = yourList.stream().sorted(Collections.reverseOrder()).collect(Collectors.toList());