Java Collections Arrays, Lists, Strings, Sets, Maps SoftUni Team Technical Trainers Software University

2 1.Arrays  int[], String[], etc. 2.Lists  ArrayList 3.Strings  String str = "Hello"; 4.Sets  HashSet, TreeSet 5.Maps  HashMap, TreeMap Table of Contents

3  The "Java Fundamentals" course is NOT for absolute beginners  Take the "C# Basics" course at SoftUni first:  The course is for beginners, but with previous coding skills  Requirements  Coding skills – entry level  Computer English – entry level  Logical thinking Warning: Not for Absolute Beginners

Arrays

What are Arrays?  In programming array is a sequence of elements  All elements are of the same type  The order of the elements is fixed  Has fixed size ( length ) Array of 5 elements Element index …………… Element of an array

6  Reference data type  Variable “classmates” holds addresses in the heap as values.  Each address points to a separate value in the heap memory. How arrays are stored in the memory

7  Allocating an array of 10 integers:  Assigning values to the array elements:  Accessing array elements by index: Working with Arrays in Java int[] numbers = new int[10]; for (int i=0; i<numbers.length; i++) numbers[i] = i+1; numbers[i] = i+1; numbers[3] = 20; numbers[5] = numbers[2] + numbers[7];

8  You may define an array of any type, e.g. String : Arrays of Strings String[] names = { "Peter", "Maria", "Katya", "Todor" }; for (int i = 0; i<names.length; i++) { System.out.printf("names[%d] = %s\n", i, names[i]); System.out.printf("names[%d] = %s\n", i, names[i]);} for (String name : names) { System.out.println(name); System.out.println(name);} names[4] = "Nakov"; // ArrayIndexOutOfBoundsException names.length = 5; // array.length is read-only field

9 Read, Sort and Print Array of n Strings Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); scanner.nextLine(); String[] lines = new String[n]; for (int i = 0; i < n; i++) { lines[i] = scanner.nextLine(); lines[i] = scanner.nextLine();}Arrays.sort(lines); for (int i = 0; i < lines.length; i++) { System.out.println(lines[i]); System.out.println(lines[i]);}

Arrays Live Demo

Lists Using ArrayList

12  In Java arrays have fixed length  Cannot add / remove / insert elements  Lists are like resizable arrays  Allow add / remove / insert of elements  Lists in Java are defined through the ArrayList class  Where E is the type of the list, e.g. String or Integer Lists in Java ArrayList numbers = new ArrayList (); numbers.add(5); System.out.println(numbers.get(0)); // 5

13  Reference data type  Variable “ages” holds pointers to Objects in the heap as values.  Each Object has an address that points to a value in the memory.  Future reference  Arrays vs ArrayLists Arrays vs ArrayLists How array lists are stored in the memory

14 ArrayList – Example ArrayList names = new ArrayList () {{ add("Peter");add("Maria");add("Katya");add("Todor");}}; names.add("Nakov"); // Peter, Maria, Katya, Todor, Nakov names.remove(0); // Maria, Katya, Todor, Nakov names.remove(1); // Maria, Todor, Nakov names.remove("Todor"); // Maria, Nakov names.addAll(Arrays.asList("Alice", "Tedy")); // Maria, Nakov, Alice, Tedy // Maria, Nakov, Alice, Tedy names.add(3, "Sylvia"); // Maria, Nakov, Alice, Sylvia, Tedy names.set(2, "Mike"); // Maria, Nakov, Mike, Sylvia, Tedy System.out.println(names);

15 ArrayList – Example // This will not compile! ArrayList intArr = new ArrayList (); ArrayList nums = new ArrayList<>( Arrays.asList(5, -3, 10, 25)); Arrays.asList(5, -3, 10, 25)); nums.add(55); // 5, -3, 10, 25, 55 System.out.println(nums.get(0)); // 5 System.out.println(nums); // [5, -3, 10, 25, 55] nums.remove(2); // 5, -3, 25, 55 nums.set(0, 101); // 101, -3, 25, 55 System.out.println(nums); // [101, -3, 25, 55]

ArrayList Live Demo

Strings Basic String Operations

What Is String?  Strings are indexed sequences of Unicode characters  Represented by the String class in Java  Characters accessed by index: 0 … length()-1  Example: string s = "Hello, SoftUni!"; Hello, SoftUni!s

19  Strings in Java  Know their number of characters: length()  Can be accessed by index: charAt(0 … length()-1)  Reference types  Stored in the heap (dynamic memory)  Can have null value (missing value)  Strings cannot be modified (immutable)  Most string operations return a new String instance  StringBuilder class is used to build stings Working with Strings

20 Strings – Examples String str = "SoftUni"; System.out.println(str); for (int i = 0; i < str.length(); i++) { System.out.printf("str[%d] = %s\n", i, str.charAt(i)); System.out.printf("str[%d] = %s\n", i, str.charAt(i));} System.out.println(str.indexOf("Uni")); // 4 System.out.println(str.indexOf("uni")); // -1 (not found) System.out.println(str.substring(4, 7)); // Uni System.out.println(str.replace("Soft", "Hard")); // HardUni System.out.println(str.toLowerCase()); // softuni System.out.println(str.toUpperCase()); // SOFTUNI

21 Strings – Examples (2) String firstName = "Steve"; String lastName = "Jobs"; int age = 56; System.out.println(firstName + " " + lastName + " (age: " + age + ")"); // Steve Jobs (age: 56) " (age: " + age + ")"); // Steve Jobs (age: 56) String allLangs = "C#, Java; HTML, CSS; PHP, SQL"; String[] langs = allLangs.split("[, ;]+"); for (String lang : langs) { System.out.println(lang); System.out.println(lang);} System.out.println("Langs = " + String.join(", ", langs)); System.out.println(" \n\n Software University ".trim());

22  The == operator does not work correctly for strings!  Use String.equals(String) and String.compareTo(String) Comparing Strings in Java String[] words = "yes yes".split(" "); System.out.println("words[0] = " + words[0]); // yes System.out.println("words[1] = " + words[0]); // yes System.out.println(words[0] == words[1]); // false System.out.println(words[0].equals(words[1])); // true System.out.println("Alice".compareTo("Mike")); // < 0 System.out.println("Alice".compareTo("Alice")); // == 0 System.out.println("Mike".compareTo("Alice")); // > 0

Strings Live Demos

Sets HashSet and TreeSet

25  Sets in Java keep unique elements  Like lists but duplicated elements are stored only once  HashSet  Keeps a set of elements in a hash-tables  The elements are randomly ordered (by their hash code)  TreeSet  Keeps a set of elements in a red-black ordered search tree  The elements are ordered incrementally Sets in Java

26  Reference data type  Variable “names” holds hash indexes that point to Objects in the heap as values.  Each Object has an address that points to a value in the memory.  Future Reference  HashSet HashSet  TreeSet TreeSet How hash sets are stored in the memory

27 HashSet and TreeSet – Examples Set set = new TreeSet (); set.add("Pesho");set.add("Tosho");set.add("Pesho");set.add("Gosho");set.add("Maria");set.add("Alice");set.remove("Pesho"); System.out.println(set); // [Alice, Gosho, Maria, Tosho]

Maps

29  Maps in Java keep unique pairs  HashMap  Keeps a map of elements in a hash-table  The elements are randomly ordered (by their hash code)  TreeMap  Keeps a set of elements in a red-black ordered search tree  The elements are ordered incrementally by their key Maps in Java

30  Reference data type  Variable “phonebook” holds hash indexes that point to keys in the heap as values.  Each key points to a value in the memory. How hash maps are stored in the memory

31  Counting words occurrences in a list: HashMap – Examples String[] words = { "yes", "hi", "hello", "hi", "welcome", "yes", "yes", "welcome", "hi", "yes", "hello", "yes" }; "yes", "yes", "welcome", "hi", "yes", "hello", "yes" }; Map wordsCount = new HashMap (); for (String word : words) { Integer count = wordsCount.get(word); Integer count = wordsCount.get(word); if (count == null) { if (count == null) { count = 0; count = 0; } wordsCount.put(word, count+1); wordsCount.put(word, count+1);} System.out.println(wordsCount); // {hi=3, yes=5, hello=2, welcome=2}

32  Students and their grades TreeMap – Examples HashMap > grades = new HashMap<>(); grades.put("Peter", new ArrayList<>(Arrays.asList(5))); grades.put("George", new ArrayList<>(Arrays.asList(5, 5, 6))); grades.put("Maria", new ArrayList<>(Arrays.asList(5, 4, 4))); grades.get("Peter").add(6);grades.get("George").add(6); for (String key : grades.keySet()) { System.out.println("" + key + " -> " + grades.get(key)); System.out.println("" + key + " -> " + grades.get(key));}

33  Java Oracle documentation  The book “Java Generics and Collections” Future References

Collection Querying and Traversing Iterative and Functional

Iterative Approach

36  Native traversing can be done only on collections implementing the Iterable interface  Set, Vector, List, Queue and AbstractCollection Collection Querying and Traversing <> names = ArrayList<>() (name : names) {..(name) } List names = new ArrayList<>(); for (String name : names) { System.out.println(name); }

37  Traversing cannot be done upon a Map instance  The type in a Map instance is always unique (a map cannot have duplicated keys)  This allows to traverse the keys through a collection that does not allow duplicates ( Set )  Method keySet() from the Map interface returns a Set Collection Querying and Traversing (1)

38  The type in a Map instance allows duplicates.  This allows to traverse the keys through an abstract collection ( Collection )  Method values() from the Map interface returns an abstract collection Collection Querying and Traversing (2)

39  To traverse the Key and Value pair, Java exposes a special interface called Entry similar to C#’s KeyValuePair Collection Querying and Traversing (3)

40  The pair K,V is always unique, because of the uniqueness of the keyset.  Thus the collection that holds Key-Value pair does not allow duplicates ( Set )  To retrieve a collection of Key-Value pairs, the Map interface exposes a method entrySet() which holds a Set of Entry instances Collection Querying and Traversing (4)

41 Collection Querying and Traversing (5)

42  To query a collection one needs to iterate over it and find the desired result. Collection Querying and Traversing (6) <> names = ArrayList<>() names.() names.() desiredName = (name : names) { (name.() { desiredName = name } } List names = new ArrayList<>(); names.add("gosho"); names.add("pesho"); String desiredName = null; for (String name : names) { if (name.equals("pesho") { desiredName = name; break; } }

Functional Approach

44  Querying a collection is also possible in a functional way  Methods are chained returning a new query instance  A terminal method is executed at the end  This is all possible via the Stream API available from Java 8 Collection Querying and Traversing (7)

45  Intermediate methods  distinct() – removes non-unique elements  filter(Predicate ) – filters elements (Where in LINQ)  flatMap(Function ) – transforms one Stream to another Stream. May contain different type of elements  limit(long) – limits the elements in the new Stream  map(Function ) – flatMap() without different types. Same as Select in LINQ  sorted(Comparator?) – sorts the elements in the Stream Collection Querying and Traversing (8)

46  Terminal methods  allMatch(Predicate ) – checks whether all elements in the Stream meets the predicate criteria (boolean)  anyMatch( ) – checks whether at least one element in the Stream meets the predicate criteria (boolean)  collect(Collector ) – converts a Stream to a materialized collection (List, Map, Set…)  findAny() – returns an element from the Stream. Returns Optional (same as Nullable in C#) Collection Querying and Traversing (9)

47  Terminal methods (1)  findFirst() – returns the first element from the Stream  forEach(Consumer ) – executes the consumer implementation upon each element. Void one.  forEachOrdered(Consumer ) – same as above but the elements are ordered. Not thread-safe  max(Comparator ) – returns the maximum element by a given criteria wrapped in Optional Collection Querying and Traversing (9.1)

48 Collection Querying and Traversing (10) <> names = ArrayList<>() names.().(n -> n.() > ).(.::) <> first = names.().()..(first.()) List names = new ArrayList<>(); names.stream().filter(n -> n.length() > 8).forEach(System.out::println); Optional first = names.stream().findFirst(); System.out.println(first.get());

49 Collection Querying and Traversing (11) > venues = <>() venues.().().(entry -> { entry.().().().((innerEntry1innerEntry2) -> {.(innerEntry1.()innerEntry2.()) }).(innerEntry -> {..(innerEntry.())..()..(innerEntry.()) }) }) LinkedHashMap > venues = new LinkedHashMap<>(); venues.entrySet().stream().forEach(entry -> { entry.getValue().entrySet().stream().sorted((innerEntry1, innerEntry2) -> { return Integer.compare(innerEntry1.getValue(), innerEntry2.getValue()); }).forEach(innerEntry -> { System.out.println(innerEntry.getKey()); System.out.println(" "); System.out.println(innerEntry.getValue()); }); });

50  Monads with Java 8 Stream (Bulgarian) Future References

Recursion

52  In order to understand recursion one first has to understand recursion  Recursion is the process of repeating items in self-similar way  Method calling itself until bottom is reached  Recursion uses the system stack Recursion

53  Recursion is highly used in traversal algorithms Recursion (1)

54  Traversing file system  Each object of type File is a file unless List is populated  Then it’s folder  This folder can have child folders with files too Recursion (2)

55 Recursion (3) ([] args) { diskC = () diskC.= autoexec = () autoexec.= diskC..(autoexec) errorLog = () errorLog.= diskC..(errorLog) } public static void main(String[] args) { File diskC = new File(); diskC.name = "Hard Drive (C)"; File autoexec = new File(); autoexec.name = "autoexec.bat"; // just a file diskC.files.add(autoexec); // autoexec.bat is now child of disk C File errorLog = new File(); errorLog.name = "errors.log"; // just a file diskC.files.add(errorLog); // error.log is now child of disk C }

56  It’s easy to retrieve Disk C’s children  But what if one of the files is a folder? Recursion (4) (f : diskC.) {..(f.) } for (File f : diskC.files) { System.out.println(f.name); } windows = () windows.= system32 = () system32.= windows..(system32) diskC..(windows) File windows = new File(); windows.name = "Windows"; File system32 = new File(); system32.name = "System 32"; windows.files.add(system32); // windows is a folder now diskC.files.add(windows); // windows folder is child of disk C

57  The foreach has to be called upon each child Recursion (5)

58  Starting from Disk C until reaching the last file  Output: Recursion (6)

59  Arrays, Strings and Collections: 1. Arrays: int[], String[], etc. 2. Strings: String str = "Hello"; 3. Lists: ArrayList 4. Sets: HashSet, TreeSet 5. Maps: HashMap, TreeMap Summary

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61  This course (slides, examples, demos, videos, homework, etc.) is licensed under the "Creative Commons Attribution- NonCommercial-ShareAlike 4.0 International" licenseCreative Commons Attribution- NonCommercial-ShareAlike 4.0 International  Attribution: this work may contain portions from  "Fundamentals of Computer Programming with Java" book by Svetlin Nakov & Co. under CC-BY-SA licenseFundamentals of Computer Programming with JavaCC-BY-SA  "C# Basics" course by Software University under CC-BY-NC-SA licenseC# BasicsCC-BY-NC-SA License

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