我們今天來討論比較器 & 集合的用法

利用比較器來實現排序

List<String> sampleStrings =
    Arrays.asList("this", "is", "a", "list", "of", "strings");
Collections.sort(sampleStrings);
System.out.println(sampleStrings);

List<String> result = sampleStrings.stream()
                                .sorted()
                                .collect(Collectors.toList());
System.out.println(result);
//output:
//[a, is, list, of, strings, this]
//[a, is, list, of, strings, this]

//Lambda expression
List<String> result = sampleStrings.stream()
        .sorted((s1, s2) -> s1.length() - s2.length())
        .collect(toList());

//Comparator.comparingInt static method reference
List<String> result = sampleStrings.stream()
    .sorted(Comparator.comparingInt(String::length))
    .collect(toList());

//using Comparator static method
sampleStrings.stream()
        .sorted(Comparator.comparing(String::length)
        .thenComparing(Comparator.naturalOrder()))
        .collect(toList());

串流轉成 Collection

此為串流聚合操作，利用 Collectors.toList, toSet, toCollection 的靜態方法
這我們之前常用，就不多舉例了

List<String> superHeroes =
    Stream.of("Mr. Furious", "The Blue Raja", "The Shoveler",
              "The Bowler", "Invisible Boy", "The Spleen", "The Sphinx")
          .collect(Collectors.toList());

將線性集合加入 Map

我們想要把一個線性集合加入到 Map，key 是使用物件的其中一個特性，可以使用 Collectors.toMap，具體例子如下：

class Book {
    private int id;
    private String name;
    private double price;
}

List<Book> books = Arrays.asList(
    new Book(1, "Modern Java Recipes", 49.99),
    new Book(2, "Java 8 in Action", 49.99),
    new Book(3, "Java SE8 for the Really Impatient", 39.99),
    new Book(4, "Functional Programming in Java", 27.64),
    new Book(5, "Making Java Groovy", 45.99)
    new Book(6, "Gradle Recipes for Android", 23.76)
);

Map<Integer, Book> bookMap = books.stream()
    .collect(Collectors.toMap(Book::getId, b -> b));
//or
Map<Integer, Book> bookMap = books.stream()
    .collect(Collectors.toMap(Book::getId, Function.identity()));

System.out.println(bookMap);

參考資料：

現代 Java - 輕鬆解決 Java 8 與 9 的難題（O'REILLY）