1. Översikt
Den här artikeln kommer att illustrera hur man använder sortering på Array , List , Set och Map i Java 7 och Java 8.
2. Sortera med matris
Låt oss börja med att sortera heltalmatriser först med metoden Arrays.sort () .
Vi definierar följande int- arrays i en @Before jUnit-metod:
@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. Sortering komplett matris
Låt oss nu använda det enkla Array.sort () API:
@Test public void givenIntArray_whenUsingSort_thenSortedArray() { Arrays.sort(toSort); assertTrue(Arrays.equals(toSort, sortedInts)); }Den osorterade matrisen är nu helt sorterad:
[1, 5, 7, 66, 88, 89, 123, 200, 255]Som nämnts i den officiella JavaDoc använder Arrays.sort dubbelsidig Quicksort på primitiver . Det erbjuder O (n log (n)) prestanda och är vanligtvis snabbare än traditionella (en-pivot) Quicksort-implementeringar. Den använder dock en stabil, adaptiv, iterativ implementering av mergesort-algoritm för Array of Objects.
2.2. Sortera del av en matris
Arrays.sort har ytterligare en slags API: er - som vi diskuterar här:
Arrays.sort(int[] a, int fromIndex, int toIndex)Detta sorterar bara en del av matrisen mellan de två indexen.
Låt oss ta en titt på ett snabbt exempel:
@Test public void givenIntArray_whenUsingRangeSort_thenRangeSortedArray() { Arrays.sort(toSort, 3, 7); assertTrue(Arrays.equals(toSort, sortedRangeInts)); }Sorteringen görs endast på följande undergruppselement ( toIndex skulle vara exklusivt):
[255, 7, 88, 200]Den resulterande sorterade undergruppen inklusive huvudmatrisen skulle vara:
[5, 1, 89, 7, 88, 200, 255, 123, 66]2.3. Java 8 Arrays.sort vs Arrays.parallelSort
Java 8 levereras med ett nytt API - parallelSort - med en liknande signatur som Arrays.sort () API:
@Test public void givenIntArray_whenUsingParallelSort_thenArraySorted() { Arrays.parallelSort(toSort); assertTrue(Arrays.equals(toSort, sortedInts)); }Bakom kulisserna för parallelSort () bryter den upp matrisen i olika undermatriser (enligt granularitet i algoritmen för parallelSort ). Varje undergrupp sorteras med Arrays.sort () i olika trådar så att sorteringen kan exekveras parallellt och slås samman slutligen som en sorterad matris.
Observera att den gemensamma poolen ForJoin används för att utföra dessa parallella uppgifter och sedan slå samman resultaten.
Resultatet av Arrays.parallelSort kommer att bli detsamma som Array. Sort naturligtvis, det är bara en fråga om att utnyttja multi-threading.
Slutligen finns det liknande varianter av API Arrays.sort i Arrays.parallelSort också:
Arrays.parallelSort (int [] a, int fromIndex, int toIndex);3. Sortera en lista
Låt oss nu använda API: n för Collections.sort () i java.utils.Collections - för att sortera en lista över heltal:
@Test public void givenList_whenUsingSort_thenSortedList() { List toSortList = Ints.asList(toSort); Collections.sort(toSortList); assertTrue(Arrays.equals(toSortList.toArray(), ArrayUtils.toObject(sortedInts))); }Den lista före sortering kommer att innehålla följande:
[5, 1, 89, 255, 7, 88, 200, 123, 66]Och naturligtvis efter sortering:
[1, 5, 7, 66, 88, 89, 123, 200, 255]Som nämnts i Oracle JavaDoc for Collections.Sort använder den en modifierad sammanslagningssort och erbjuder garanterad n log (n) -prestanda.
4. Sortera en uppsättning
Låt oss sedan använda Collections.sort () för att sortera en LinkedHashSet .
Vi använder LinkedHashSet eftersom det bibehåller införingsordningen.
Lägg märke till hur, i syfte att använda sorterings API i samlingar - vi första omslag uppsättningen i en lista :
@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())); }Den Comparator.reverseOrder () metoden reverserar ordningen som infördes genom den naturliga ordningen.
5. Sorterings Karta
I det här avsnittet börjar vi titta på att sortera en karta - både efter nycklar och efter värden.
Låt oss först definiera kartan vi ska sortera:
@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. Sortera karta efter nycklar
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:
- in the natural order (Using the Comparable Interface) or
- 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.
Vi såg också en kort introduktion om hur funktioner i Java 8 kan vara användbara vid sortering som användning av Lambdas, jämförelse () och sedanJämförelse () och parallelSort () .
Alla exempel som används i artikeln finns på GitHub.