How to sort an ArrayList?

问题:

I have a List of doubles in java and I want to sort ArrayList in descending order.

Input ArrayList is as below:

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);

The out put should be like this

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

回答1:

Collections.sort(testList);
Collections.reverse(testList);

That will do what you want. Remember to import Collections though!

Here is the documentation for Collections.

回答2:

Descending:

Collections.sort(mArrayList, new Comparator<CustomData>() {
    @Override
    public int compare(CustomData lhs, CustomData rhs) {
        // -1 - less than, 1 - greater than, 0 - equal, all inversed for descending
        return lhs.customInt > rhs.customInt ? -1 : (lhs.customInt < rhs.customInt) ? 1 : 0;
    }
});

回答3:

Use util method of java.util.Collections class, i.e

Collections.sort(list)

In fact, if you want to sort custom object you can use

Collections.sort(List<T> list, Comparator<? super T> c) 

see collections api

回答4:

For your example, this will do the magic in Java 8

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

But if you want to sort by some of the fields of the object you are sorting, you can do it easily by:

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());

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

回答5:

Using lambdas (Java8), and stripping it down to the barest of syntax (the JVM will infer plenty in this case), you get:

Collections.sort(testList, (a, b) -> b.compareTo(a));

A more verbose version:

// Implement a reverse-order Comparator by lambda function
Comparator<Double> comp = (Double a, Double b) -> {
    return b.compareTo(a);
};

Collections.sort(testList, comp);

The use of a lambda is possible because the Comparator interface has only a single method to implement, so the VM can infer which method is implementing. Since the types of the params can be inferred, they don\'t need to be stated (i.e. (a, b) instead of (Double a, Double b). And since the lambda body has only a single line, and the method is expected to return a value, the return is inferred and the braces aren\'t necessary.

回答6:

With Java8 there is a default sort method on the List interface that will allow you to sort the collection if you provide a Comparator. You can easily sort the example in the question as follows:

testList.sort((a, b) -> Double.compare(b, a));

Note: the args in the lambda are swapped when passed in to Double.compare to ensure the sort is descending

回答7:

You can use Collections.sort(list) to sort list if your list contains Comparable elements. Otherwise I would recommend you to implement that interface like here:

public class Circle implements Comparable<Circle> {}

and of course provide your own realization of compareTo method like here:

@Override
    public int compareTo(Circle another) {
        if (this.getD()<another.getD()){
            return -1;
        }else{
            return 1;
        }
    }

And then you can again use Colection.sort(list) as now list contains objects of Comparable type and can be sorted. Order depends on compareTo method. Check this https://docs.oracle.com/javase/tutorial/collections/interfaces/order.html for more detailed information.

回答8:

Collections.sort allows you to pass an instance of a Comparator which defines the sorting logic. So instead of sorting the list in natural order and then reversing it, one can simply pass Collections.reverseOrder() to sort in order to sort the list in reverse order:

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

As mentioned by @Marco13, apart from being more idiomatic (and possibly more efficient), using the reverse order comparator makes sure that the sort is stable (meaning that the order of elements will not be changed when they are equal according to the comparator, whereas reversing will change the order)

回答9:

//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;

}
}

回答10:

if you are using Java SE 8, then this might be of help.

//create a comparator object using a Lambda expression
Comparator<Double> compareDouble = (d1, d2) -> d1.compareTo(d2);

//Sort the Collection in this case \'testList\' in reverse order
Collections.sort(testList, Collections.reverseOrder(compareDouble));

//print the sorted list using method reference only applicable in SE 8
testList.forEach(System.out::println);

回答11:

|*| Sorting an List :

import java.util.Collections;

|=> Sort Asc Order :

Collections.sort(NamAryVar);

|=> Sort Dsc Order :

Collections.sort(NamAryVar, Collections.reverseOrder());

---

|*| Reverse the order of List :

Collections.reverse(NamAryVar);

回答12:

you can do like this:

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

Collection has a default Comparator that can help you with that.

Also, if you want to use some Java 8 new features, you can do like that:

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

回答13:

You can use like that

              ArrayList<Group> groupList = new ArrayList<>();
              Collections.sort(groupList, Collections.reverseOrder());
              Collections.reverse(groupList);

回答14:

For example I have a class Person: String name, int age ==>Constructor 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]]

}

回答15:

With Eclipse Collections you could create a primitive double list, sort it and then reverse it to put it in descending order. This approach would avoid boxing the doubles.

MutableDoubleList doubleList =
    DoubleLists.mutable.with(
        0.5, 0.2, 0.9, 0.1, 0.1, 0.1, 0.54, 0.71,
        0.71, 0.71, 0.92, 0.12, 0.65, 0.34, 0.62)
        .sortThis().reverseThis();
doubleList.each(System.out::println);

If you want a List<Double>, then the following would work.

List<Double> objectList =
    Lists.mutable.with(
        0.5, 0.2, 0.9, 0.1, 0.1, 0.1, 0.54, 0.71,
        0.71, 0.71, 0.92, 0.12, 0.65, 0.34, 0.62)
        .sortThis(Collections.reverseOrder());
objectList.forEach(System.out::println);

If you want to keep the type as ArrayList<Double>, you can initialize and sort the list using the ArrayListIterate utility class as follows:

ArrayList<Double> arrayList =
    ArrayListIterate.sortThis(
            new ArrayList<>(objectList), Collections.reverseOrder());
arrayList.forEach(System.out::println);

Note: I am a committer for Eclipse Collections.

回答16:

In JAVA 8 its much easy now.

List<String> alphaNumbers = Arrays.asList(\"one\", \"two\", \"three\", \"four\");
List<String> alphaNumbersUpperCase = alphaNumbers.stream()
    .map(String::toUpperCase)
    .sorted()
    .collect(Collectors.toList());
System.out.println(alphaNumbersUpperCase); // [FOUR, ONE, THREE, TWO]

-- For reverse use this

.sorted(Comparator.reverseOrder())

回答17:

The following line should do the thick

testList.sort(Collections.reverseOrder());

回答18:

Just use loop or default sort() function.

for(int n = 0; n<= arrList[i]; n++){
   i