在Java中非常大的集合或列表中搜索与regex匹配的字符串

Question

我的所有单词的文件大约是60MB，搜索押韵现在需要大约几分钟。

SELECT *
FROM table 
WHERE word LIKE '%rhyme'

多谢帮忙！

package pl.kamilkoszykowski.dopewriter;
import org.springframework.web.bind.annotation.*;
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.*;
import java.util.concurrent.*;

@RestController
@CrossOrigin("http://localhost:4200")
public class Controller {

    Set<String> dictionary = getDictionary(); // SET WITH WORDS

    @GetMapping("/rhyme/{word}")
    public Set<String> rhymes(@PathVariable String word) throws InterruptedException {

        String regex = "\\b[A-Za-z]*" + word + "\\b"; // RHYME TO SEARCH FOR IN DICTIONARY SET

        Set<String> rhymesList = new HashSet<>();

        int numRunnables = 64;
        BlockingQueue<Runnable> queue = new ArrayBlockingQueue<Runnable>(numRunnables, true);
        RejectedExecutionHandler handler = new ThreadPoolExecutor.CallerRunsPolicy();
        ExecutorService executor = new ThreadPoolExecutor(numRunnables, numRunnables, 0L, TimeUnit.MILLISECONDS, queue, handler);

        for (String a : dictionary) {
            executor.execute(new Runnable() {

                @Override
                public void run() {
                    if (a.matches(regex)) {
                        rhymesList.add(a);
                    }
                }
            });
        }

        executor.shutdown();
        while (executor.isTerminated() == false){
            Thread.sleep(50);
        }

        return rhymesList;
    }

    public Set<String> getDictionary() { // READING TXT FILE
        try {
            List<String> list = new ArrayList<>(List.of(Files.readString(Paths.get("src/main/resources/dictionary.txt")).split(",")));
            return new HashSet<>(list);
        } catch (IOException e) {
            return null;
        }
    }
}

Answer 1

迭代集合本身是一个耗时的过程。多种解决方式：

1. 使用java流。--与自定义线程池执行器相比，实现更快，速度更快。
2. 保持数据的排序集。做一个排序技术以更快地得到结果。仍然不够快。
3. 将数据存储在trie数据结构中，并获取与regex匹配的根下的所有单词。--推荐。