How to Sort an Array, List, Map or Stream in Java
Learn to sort a Java Set , List and Map of primitive types and custom objects using Comparator, Comparable and new lambda expressions. We will learn to sort in ascending and descending order as well.
//Sorting an array Arrays.sort( arrayOfItems ); Arrays.sort( arrayOfItems, Collections.reverseOrder() ); Arrays.sort(arrayOfItems, 2, 6); Arrays.parallelSort(arrayOfItems); //Sorting a List Collections.sort(numbersList); Collections.sort(numbersList, Collections.reverseOrder()); //Sorting a Set Set to List -> Sort -> List to Set Collections.sort(numbersList); //Sorting a Map TreeMap treeMap = new TreeMap<>(map); unsortedeMap.entrySet() .stream() .sorted(Map.Entry.comparingByValue()) .forEachOrdered(x -> sortedMap.put(x.getKey(), x.getValue())); //Java 8 Lambda Comparator nameSorter = (a, b) -> a.getName().compareToIgnoreCase(b.getName()); Collections.sort(list, nameSorter); Collections.sort(list, Comparator.comparing(Employee::getName)); //Group By Sorting Collections.sort(list, Comparator .comparing(Employee::getName) .thenComparing(Employee::getDob));1. Sorting a List of Objects
To sort a list of objects, we have two popular approaches i.e. Comparable and Comparator interfaces. In the upcoming examples, we will sort a collection of Employee objects in different ways.
public class Employee implements Comparable
Comparable interface enables the natural ordering of the classes that implements it. This makes the classes comparable to its other instances.
A class implementing Comparable interface must override compareTo() method in which it can specify the comparison logic between two instances of the same class.
- Lists (and arrays) of objects that implement Comparable interface can be sorted automatically by Collections.sort() and Arrays.sort() APIs.
- Objects that implement this interface will be automatically sorted when put in a sorted map (as keys) or sorted set (as elements).
- It is strongly recommended (though not required) that natural orderings be consistent with equals() method. Virtually all Java core classes that implement Comparable have natural orderings that are consistent with equals() .
ArrayList list = new ArrayList<>(); //add employees.. Collections.sort(list);To sort the list in reversed order, the best way is to use the Comparator.reversed() API that imposes the reverse ordering.
ArrayList list = new ArrayList<>(); //add employees.. Collections.sort(list, Comparator.reversed());When not seeking the natural ordering, we can take the help of Comparator interface to apply custom sorting behavior.
Comparator does not require modifying the source code of the class. We can create the comparison logic in a separate class which implements Comparator interface and override its compare() method.
During sorting, we pass an instance of this comparator to sort() method along with the list of custom objects.
For example, we want to sort the list of employees by their first name, while the natural sorting has been implemented by id field. So, to sort on name field, we must write the custom sorting logic using Comparator interface.
import java.util.Comparator; public class NameSorter implements Comparator<Employee> < @Override public int compare(Employee e1, Employee e2) < return e1.getName().compareToIgnoreCase( e2.getName() ); >>Notice the use of NameSorter in sort() method as the second argument in the given example.
ArrayList list = new ArrayList<>(); //add employees to list Collections.sort(list, new NameSorter());To do the reverse sorting, we just need to call the reversed() method on the Comparator instance.
ArrayList list = new ArrayList<>(); //add employees to list Collections.sort(list, new NameSorter().reversed());1.3. Sorting with Lambda Expressions
Lambda expressions help in writing Comparator implementations on the fly. We do not need to create a separate class to provide the one-time comparison logic.
Comparator nameSorter = (a, b) -> a.getName().compareToIgnoreCase(b.getName()); Collections.sort(list, nameSorter);To apply SQL style sorting on a collection of objects on different fields (group by sort), we can use multiple comparators in a chain. This chaining of comparators can be created using Comparator.comparing() and Comparator.thenComparing() methods.
For example, we can sort the list of employees by name and then sort again by their age.
ArrayList list = new ArrayList<>(); //add employees to list Collections.sort(list, Comparator .comparing(Employee::getName) .thenComparing(Employee::getDob));Use java.util.Arrays.sort() method to sort a given array in a variety of ways. The sort() is an overloaded method that takes all sorts of types as the method argument.
This method implements a Dual-Pivot Quicksort sorting algorithm that offers O(n log(n)) performance on all data sets and is typically faster than traditional (one-pivot) Quicksort implementations.
Java program to sort an array of integers in ascending order using Arrays.sort() method.
//Unsorted array Integer[] numbers = new Integer[] < 15, 11, . >; //Sort the array Arrays.sort(numbers);Java provides Collections.reverseOrder() comparator to reverse the default sorting behavior in one line. We can use this comparator to sort the array in descending order.
Note that all elements in the array must be mutually comparable by the specified comparator.
//Unsorted array Integer[] numbers = new Integer[] < 15, 11, . >; //Sort the array in reverse order Arrays.sort(numbers, Collections.reverseOrder());Arrays.sort() method is an overloaded method and takes two additional parameters i.e. fromIndex (inclusive) and toIndex (exclusive).
When provided above arguments, the array will be sorted within the provided range from position fromIndex to position toIndex .
Given below is an example to sort the array from element 9 to 18. i.e. will be sorted and the rest elements will not be touched.
//Unsorted array Integer[] numbers = new Integer[] < 15, 11, 9, 55, 47, 18, 1123, 520, 366, 420 >; //Sort the array Arrays.sort(numbers, 2, 6); //Print array to confirm System.out.println(Arrays.toString(numbers));[15, 11, 9, 18, 47, 55, 1123, 520, 366, 420]Java 8 introduced lots of new APIs for parallel processing data sets and streams. One such API is Arrays.parallelSort() .
The parallelSort() method breaks the array into multiple sub-arrays and each sub-array is sorted with Arrays.sort() in different threads. Finally, all sorted sub-arrays are merged into one sorted array.
The output of the parallelSort() and sort() , both APIs, will be same at last. It’s just a matter of leveraging the Java concurrency.
//Parallel sort complete array Arrays.parallelSort(numbers); //Parallel sort array range Arrays.parallelSort(numbers, 2, 6); //Parallel sort array in reverse order Arrays.parallelSort(numbers, Collections.reverseOrder());There is no direct support for sorting the Set in Java. To sort a Set , follow these steps:
- Convert Set to List .
- Sort List using Collections.sort() API.
- Convert List back to Set .
//Unsorted set HashSet numbersSet = new LinkedHashSet<>(); //with Set items List numbersList = new ArrayList(numbersSet) ; //set -> list //Sort the list Collections.sort(numbersList); //sorted set numbersSet = new LinkedHashSet<>(numbersList); //list -> setA Map is the collection of key-value pairs. So logically, we can sort the maps in two ways i.e. sort by key or sort by value.
The best and most effective a sort a map by keys is to add all map entries in TreeMap object. TreeMap stores the keys in sorted order, always.
HashMap map = new HashMap<>(); //Unsorted Map TreeMap treeMap = new TreeMap<>(map); //Sorted by map keysIn Java 8, Map.Entry class has static method comparingByValue() to help us in sorting the Map by values.
The comparingByValue() method returns a Comparator that compares Map.Entry in natural order on values.
HashMap unSortedMap = new HashMap<>(); //Unsorted Map //LinkedHashMap preserve the ordering of elements in which they are inserted LinkedHashMap sortedMap = new LinkedHashMap<>(); unSortedMap.entrySet() .stream() .sorted(Map.Entry.comparingByValue()) .forEachOrdered(x -> sortedMap.put(x.getKey(), x.getValue()));In the above-given examples, we learned to sort an Array, List, Map, and Set.
We saw different ways to initialize and use Comparator interface including lambda expressions. We also learned to effectively use the Comparator interface.
How do I sort an array of numbers in java? [closed]
I'm trying to sort an array and looked around and saw people using Arrays.sort() so I tired it but it keeps giving me an error. Help please!
Well yes, you're calling Arrays.sort but intArray isn't an array - it's an ArrayList . Looking at the documentation for Arrays.sort , which method did you expect it to call? "I'm trying to sort an array" - no, you're not.
What do you mean? You don't "call" an index. It's unclear what you mean, but it doesn't sound particularly related to your question.
3 Answers 3
Here is your Error: int intSortedArray = Arrays.sort(intArray); replace it by Collections.sort(intArray);
Make use of Collections utility class sort() method like this :-
public static void main(String[] args) < System.out.println("Please enter the number of values you would like to enter: "); Scanner scan = new Scanner(System.in); int intNumberOfNumbers = scan.nextInt(); ArrayList intArray = new ArrayList(); for (int i = 0; i < intNumberOfNumbers; i++) < System.out.println("Please enter a value for index " + i + ":"); int intValue = scan.nextInt(); intArray.add(intValue); >System.out.println(intArray); Collections.sort(intArray); // this will sort the Integer List System.out.println("sorted list : "+intArray); >More about Collections.sort() here.
You are facing a compilation error at the following:
int intSortedArray = Arrays.sort(intArray);Arrays.sort returns a void and hence you cannot assign it to a variable
Also the Arrays.sort and their overloads works on arrays of primitive types and hence would not be able to process ArrayList. Hence you would end up with a compilation error.
Refer to the documentation of Arrays.sort method to understand the correct arguments and the return types.
The correct way to sort then would be to use Collections.sort .