Java8Stream函数将一个字母表组合成一个列表映射

Question

LIST<T>.stream().collect(Collectors.groupingBy(POJO::GET_METHOD))

public class AnagramsGrouping {
    static class Word {
        public String original;

        public Word(String word) {
            original = word;
        }

        public String getKey() {
            char[] characters = input.toCharArray();
            Arrays.sort(characters);
            return new String(characters);
        }

        public String toString() {
            return original;
        }
    }

    public static void main(String[] args) {
        List<Word> words = Arrays.asList(new Word("pool"), new Word("loop"),
                new Word("stream"), new Word("arc"), new Word("odor"),
                new Word("car"), new Word("rood"), new Word("meats"),
                new Word("fires"), new Word("fries"), new Word("night"),
                new Word("thing"), new Word("mates"), new Word("teams"));

        Map<String, List<Word>> anagrams = words.stream().collect(
                Collectors.groupingBy(Word::getKey));

        System.out.println(anagrams);
    }
}

{door=[odor, rood], acr=[arc, car], ghint=[night, thing],
 aemrst=[stream], efirs=[fires, fries], loop=[pool, loop],
 aemst=[meats, mates, teams]}

List<String> words2 = Arrays.asList("pool", "loop", "stream", "arc",
        "odor", "car", "rood", "meats", "fires", "fries",
        "night", "thing", "mates", "teams");

words2.stream().collect(Collectors.toMap(w -> sortChars(w), w -> w));

但是这段代码会生成一个键冲突,因为它是1-1的映射。

Exception in thread "main" java.lang.IllegalStateException: Duplicate key pool

这很有道理...有没有一种方法可以将它们分组到与第一个使用groupingby解决方案相似的输出中，但不使用POJO包装值？

Answer 1

单参数groupingby收集器完全可以执行您想要的操作。它对其输入进行分类，您已经使用sortchars（或前面示例中的getkey)进行了分类。在同一键下分类的每个流值都被放入一个列表中，这是映射的值。因此，我们有：

Map<String, List<String>> anagrams =
    words2.stream().collect(Collectors.groupingBy(w -> sortChars(w)));

给出输出

{door=[odor, rood], acr=[arc, car], ghint=[night, thing], aemrst=[stream],
efirs=[fires, fries], loop=[pool, loop], aemst=[meats, mates, teams]}

您还可以使用方法引用：

Map<String, List<String>> anagrams =
    words2.stream().collect(Collectors.groupingBy(GroupingAnagrams::sortChars));

Map<String, Long> anagrams =
    words2.stream().collect(
        Collectors.groupingBy(GroupingAnagrams::sortChars, Collectors.counting()));

{door=2, acr=2, ghint=2, aemrst=1, efirs=2, loop=2, aemst=3}

public static String sortChars(String input) {
    char[] characters = input.toCharArray();
    Arrays.sort(characters);
    return new String(characters);
}