Ordinamento in Java

1. Panoramica

Questo articolo illustrerà come applicare l'ordinamento a Array , List , Set e Map in Java 7 e Java 8.

2. Ordinamento con array

Iniziamo ordinando prima gli array di interi utilizzando il metodo Arrays.sort () .

Definiremo i seguenti array int in un metodo @Before jUnit:

@Before public void initVariables () { toSort = new int[] { 5, 1, 89, 255, 7, 88, 200, 123, 66 }; sortedInts = new int[] {1, 5, 7, 66, 88, 89, 123, 200, 255}; sortedRangeInts = new int[] {5, 1, 89, 7, 88, 200, 255, 123, 66}; ... }

2.1. Ordinamento dell'array completo

Usiamo ora la semplice API Array.sort () :

@Test public void givenIntArray_whenUsingSort_thenSortedArray() { Arrays.sort(toSort); assertTrue(Arrays.equals(toSort, sortedInts)); }

L'array non ordinato è ora completamente ordinato:

[1, 5, 7, 66, 88, 89, 123, 200, 255]

Come menzionato nel JavaDoc ufficiale, Arrays.sort utilizza Quicksort dual-pivot sulle primitive . Offre prestazioni O (n log (n)) ed è in genere più veloce delle tradizionali implementazioni Quicksort (one-pivot). Tuttavia, utilizza un'implementazione stabile, adattiva e iterativa dell'algoritmo Mergesort per Array of Objects.

2.2. Ordinamento di una parte di un array

Arrays.sort ha un'altra API di ordinamento , di cui parleremo qui:

Arrays.sort(int[] a, int fromIndex, int toIndex)

Questo ordinerà solo una parte dell'array, tra i due indici.

Diamo un'occhiata a un rapido esempio:

@Test public void givenIntArray_whenUsingRangeSort_thenRangeSortedArray() { Arrays.sort(toSort, 3, 7); assertTrue(Arrays.equals(toSort, sortedRangeInts)); }

L'ordinamento verrà eseguito solo sui seguenti elementi del sotto-array ( toIndex sarebbe esclusivo):

[255, 7, 88, 200]

Il sotto-array ordinato risultante incluso con l'array principale sarebbe:

[5, 1, 89, 7, 88, 200, 255, 123, 66]

2.3. Java 8 Arrays.sort contro Arrays.parallelSort

Java 8 viene fornito con una nuova API - parallelSort - con una firma simile all'API Arrays.sort () :

@Test public void givenIntArray_whenUsingParallelSort_thenArraySorted() { Arrays.parallelSort(toSort); assertTrue(Arrays.equals(toSort, sortedInts)); }

Dietro le quinte di parallelSort (), suddivide l'array in diversi sotto-array (secondo la granularità nell'algoritmo di parallelSort ). Ogni sottoarray viene ordinato con Arrays.sort () in thread diversi in modo che l' ordinamento possa essere eseguito in modo parallelo e infine unito come un array ordinato.

Notare che il pool comune ForJoin viene utilizzato per eseguire queste attività parallele e quindi unire i risultati.

Il risultato di Arrays.parallelSort sarà lo stesso di Array.sort , ovviamente, è solo questione di sfruttare il multi-threading.

Infine, ci sono varianti simili di API Arrays.sort anche in Arrays.parallelSort :

Arrays.parallelSort (int [] a, int fromIndex, int toIndex);

3. Ordinamento di un elenco

Usiamo ora l' API Collections.sort () in java.utils.Collections - per ordinare un elenco di numeri interi:

@Test public void givenList_whenUsingSort_thenSortedList() { List toSortList = Ints.asList(toSort); Collections.sort(toSortList); assertTrue(Arrays.equals(toSortList.toArray(), ArrayUtils.toObject(sortedInts))); }

L' elenco prima dell'ordinamento conterrà i seguenti elementi:

[5, 1, 89, 255, 7, 88, 200, 123, 66]

E naturalmente, dopo l'ordinamento:

[1, 5, 7, 66, 88, 89, 123, 200, 255]

Come menzionato in Oracle JavaDoc for Collections.Sort , utilizza un mergesort modificato e offre prestazioni n log (n) garantite .

4. Sorting a Set

Next, let's use Collections.sort() to sort a LinkedHashSet.

We're using the LinkedHashSet because it maintains insertion order.

Notice how, in order to use the sort API in Collectionswe're first wrapping the set in a list:

@Test public void givenSet_whenUsingSort_thenSortedSet() { Set integersSet = new LinkedHashSet(Ints.asList(toSort)); Set descSortedIntegersSet = new LinkedHashSet( Arrays.asList(new Integer[] {255, 200, 123, 89, 88, 66, 7, 5, 1})); List list = new ArrayList(integersSet); Collections.sort(Comparator.reverseOrder()); integersSet = new LinkedHashSet(list); assertTrue(Arrays.equals( integersSet.toArray(), descSortedIntegersSet.toArray())); }

The Comparator.reverseOrder() method reverses the ordering imposed by the natural ordering.

5. Sorting Map

In this section, we'll start looking at sorting a Map – both by keys and by values.

Let's first define the map we'll be sorting:

@Before public void initVariables () { .... HashMap map = new HashMap(); map.put(55, "John"); map.put(22, "Apple"); map.put(66, "Earl"); map.put(77, "Pearl"); map.put(12, "George"); map.put(6, "Rocky"); .... }

5.1. Sorting Map by Keys

We'll now extract keys and values entries from the HashMap and sort it based on the values of the keys in this example:

@Test public void givenMap_whenSortingByKeys_thenSortedMap() { Integer[] sortedKeys = new Integer[] { 6, 12, 22, 55, 66, 77 }; List
    
      entries = new ArrayList(map.entrySet()); Collections.sort(entries, new Comparator
     
      () { @Override public int compare( Entry o1, Entry o2) { return o1.getKey().compareTo(o2.getKey()); } }); Map sortedMap = new LinkedHashMap(); for (Map.Entry entry : entries) { sortedMap.put(entry.getKey(), entry.getValue()); } assertTrue(Arrays.equals(sortedMap.keySet().toArray(), sortedKeys)); }
     
    

Note how we used the LinkedHashMap while copying the sorted Entries based on keys (because HashSet doesn't guarantee the order of keys).

The Map before sorting :

[Key: 66 , Value: Earl] [Key: 22 , Value: Apple] [Key: 6 , Value: Rocky] [Key: 55 , Value: John] [Key: 12 , Value: George] [Key: 77 , Value: Pearl]

The Map after sorting by keys:

[Key: 6 , Value: Rocky] [Key: 12 , Value: George] [Key: 22 , Value: Apple] [Key: 55 , Value: John] [Key: 66 , Value: Earl] [Key: 77 , Value: Pearl] 

5.2. Sorting Map by Values

Here we will be comparing values of HashMap entries for sorting based on values of HashMap:

@Test public void givenMap_whenSortingByValues_thenSortedMap() { String[] sortedValues = new String[] { "Apple", "Earl", "George", "John", "Pearl", "Rocky" }; List
    
      entries = new ArrayList(map.entrySet()); Collections.sort(entries, new Comparator
     
      () { @Override public int compare( Entry o1, Entry o2) { return o1.getValue().compareTo(o2.getValue()); } }); Map sortedMap = new LinkedHashMap(); for (Map.Entry entry : entries) { sortedMap.put(entry.getKey(), entry.getValue()); } assertTrue(Arrays.equals(sortedMap.values().toArray(), sortedValues)); }
     
    

The Map before sorting:

[Key: 66 , Value: Earl] [Key: 22 , Value: Apple] [Key: 6 , Value: Rocky] [Key: 55 , Value: John] [Key: 12 , Value: George] [Key: 77 , Value: Pearl]

The Map after sorting by values:

[Key: 22 , Value: Apple] [Key: 66 , Value: Earl] [Key: 12 , Value: George] [Key: 55 , Value: John] [Key: 77 , Value: Pearl] [Key: 6 , Value: Rocky]

6. Sorting Custom Objects

Let's now work with a custom object:

public class Employee implements Comparable { private String name; private int age; private double salary; public Employee(String name, int age, double salary) { ... } // standard getters, setters and toString }

We'll be using the following Employee Array for sorting example in the following sections:

@Before public void initVariables () { .... employees = new Employee[] { new Employee("John", 23, 5000), new Employee("Steve", 26, 6000), new Employee("Frank", 33, 7000), new Employee("Earl", 43, 10000), new Employee("Jessica", 23, 4000), new Employee("Pearl", 33, 6000)}; employeesSorted = new Employee[] { new Employee("Earl", 43, 10000), new Employee("Frank", 33, 70000), new Employee("Jessica", 23, 4000), new Employee("John", 23, 5000), new Employee("Pearl", 33, 4000), new Employee("Steve", 26, 6000)}; employeesSortedByAge = new Employee[] { new Employee("John", 23, 5000), new Employee("Jessica", 23, 4000), new Employee("Steve", 26, 6000), new Employee("Frank", 33, 70000), new Employee("Pearl", 33, 4000), new Employee("Earl", 43, 10000)}; }

We can sort arrays or collections of custom objects either:

  1. in the natural order (Using the Comparable Interface) or
  2. in the order provided by a ComparatorInterface

6.1. Using Comparable

The natural order in java means an order in which primitive or Object should be orderly sorted in a given array or collection.

Both java.util.Arrays and java.util.Collections have a sort() method, and It's highly recommended that natural orders should be consistent with the semantics of equals.

In this example, we will consider employees with the same name as equal:

@Test public void givenEmpArray_SortEmpArray_thenSortedArrayinNaturalOrder() { Arrays.sort(employees); assertTrue(Arrays.equals(employees, employeesSorted)); }

You can define the natural order for elements by implementing a Comparable interface which has compareTo() method for comparing current object and object passed as an argument.

To understand this clearly, let's see an example Employee class which implements Comparable Interface:

public class Employee implements Comparable { ... @Override public boolean equals(Object obj) { return ((Employee) obj).getName().equals(getName()); } @Override public int compareTo(Object o) { Employee e = (Employee) o; return getName().compareTo(e.getName()); } }

Generally, the logic for comparison will be written the method compareTo. Here we are comparing the employee order or name of the employee field. Two employees will be equal if they have the same name.

Now when Arrays.sort(employees); is called in the above code, we now know what is the logic and order which goes in sorting the employees as per the age :

[("Earl", 43, 10000),("Frank", 33, 70000), ("Jessica", 23, 4000), ("John", 23, 5000),("Pearl", 33, 4000), ("Steve", 26, 6000)]

We can see the array is sorted by name of the employee – which now becomes a natural order for Employee Class.

6.2. Using Comparator

Now, let's sort the elements using a Comparator interface implementation – where we pass the anonymous inner class on-the-fly to the Arrays.sort() API:

@Test public void givenIntegerArray_whenUsingSort_thenSortedArray() { Integer [] integers = ArrayUtils.toObject(toSort); Arrays.sort(integers, new Comparator() { @Override public int compare(Integer a, Integer b) { return Integer.compare(a, b); } }); assertTrue(Arrays.equals(integers, ArrayUtils.toObject(sortedInts))); }

Now lets sort employees based on salary – and pass in another comparator implementation:

Arrays.sort(employees, new Comparator() { @Override public int compare(Employee o1, Employee o2) { return Double.compare(o1.getSalary(), o2.getSalary()); } });

The sorted Employees arrays based on salary will be:

[(Jessica,23,4000.0), (John,23,5000.0), (Pearl,33,6000.0), (Steve,26,6000.0), (Frank,33,7000.0), (Earl,43,10000.0)] 

Note that we can use Collections.sort() in a similar fashion to sort List and Set of Objects in Natural or Custom order as described above for Arrays.

7. Sorting With Lambdas

Start with Java 8, we can use Lambdas to implement the Comparator Functional Interface.

You can have a look at the Lambdas in Java 8 writeup to brush up on the syntax.

Let's replace the old comparator:

Comparator c = new Comparator() { @Override public int compare(Integer a, Integer b) { return Integer.compare(a, b); } }

With the equivalent implementation, using Lambda expression:

Comparator c = (a, b) -> Integer.compare(a, b);

Finally, let's write the test:

@Test public void givenArray_whenUsingSortWithLambdas_thenSortedArray() { Integer [] integersToSort = ArrayUtils.toObject(toSort); Arrays.sort(integersToSort, (a, b) -> { return Integer.compare(a, b); }); assertTrue(Arrays.equals(integersToSort, ArrayUtils.toObject(sortedInts))); }

As you can see, a much cleaner and more concise logic here.

8. Using Comparator.comparing and Comparator.thenComparing

Java 8 comes with two new APIs useful for sorting – comparing() and thenComparing() in the Comparator interface.

These are quite handy for the chaining of multiple conditions of the Comparator.

Let's consider a scenario where we may want to compare Employee by age and then by name:

@Test public void givenArrayObjects_whenUsingComparing_thenSortedArrayObjects() { List employeesList = Arrays.asList(employees); employees.sort(Comparator.comparing(Employee::getAge)); assertTrue(Arrays.toString(employees.toArray()) .equals(sortedArrayString)); }

In this example, Employee::getAge is the sorting key for Comparator interface implementing a functional interface with compare function.

Here's the array of Employees after sorting:

[(John,23,5000.0), (Jessica,23,4000.0), (Steve,26,6000.0), (Frank,33,7000.0), (Pearl,33,6000.0), (Earl,43,10000.0)]

Here the employees are sorted based on age.

We can see John and Jessica are of same age – which means that the order logic should now take their names into account- which we can achieve with thenComparing():

... employees.sort(Comparator.comparing(Employee::getAge) .thenComparing(Employee::getName)); ... 

After sorting with above code snippet, the elements in employee array would be sorted as:

[(Jessica,23,4000.0), (John,23,5000.0), (Steve,26,6000.0), (Frank,33,7000.0), (Pearl,33,6000.0), (Earl,43,10000.0) ]

Thus comparing() and thenComparing() definitely make more complex sorting scenarios a lot cleaner to implement.

9. Conclusion

In this article, we saw how we can apply sorting to Array, List, Set, and Map.

Abbiamo anche visto una breve introduzione su come le funzionalità di Java 8 potrebbero essere utili nell'ordinamento come l'utilizzo di Lambdas, comparing () e quindiComparing () e parallelSort () .

Tutti gli esempi utilizzati nell'articolo sono disponibili su GitHub.