HashMap 常見應用:實現 SQL JOIN

在個人上一篇文章中,講到了我本身初步認識 HashMap 的一個經驗分享:HashMap 淺析 —— LeetCode Two Sum 刷題總結。做爲一個 CRUD 工程師,平時不多接觸到基礎組件的涉及,那麼是否是很難有機會用到 HashMap 呢?java

今天,就舉一個常見的查詢例子,來看看咱們如何使用 HashMap 來提升代碼的效率。算法

已知一個 Student 類:編程

public class Student {
    private Long id;

    private String name;

    public Student(Long id, String name) {
        this.id = id;
        this.name = name;
    }

    // ---Getters And Setters---
}

和一個 Score 類:segmentfault

public class Score {
    private Long studentId;

    private String mathScore;

    private String englishScore;

    public Score(Long studentId, String mathScore, String englishScore) {
        this.studentId = studentId;
        this.mathScore = mathScore;
        this.englishScore = englishScore;
    }

    // ---Getters And Setters---
}

咱們須要把 Student 和 Score 合併到一塊兒,即類 Report:微服務

public class Report {
    private Long studentId;

    private String studentName;

    private String mathScore;

    private String englishScore;

    public Report(Long studentId, String studentName, String mathScore, String englishScore) {
        this.studentId = studentId;
        this.studentName = studentName;
        this.mathScore = mathScore;
        this.englishScore = englishScore;
    } 
}

看類的屬性咱們就明白了,這裏其實至關於在 Student 和 Score 之間作一個 Join,獲得 Report。這是咱們在編程中常見的場景(例如查詢了訂單中心,用戶中心,支付中心,合併各個中心返回的結果造成一個表單,由於各個中心是獨立的微服務,沒法使用 SQL JOIN)。測試

現有兩個 List:List<Student>List<Score>this

List<Student> students = Arrays.asList(
        new Student(1L, "Angle"),
        new Student(2L, "Baby")
);

List<Score> scores = Arrays.asList(
        new Score(1L, "90", "87"),
        new Score(2L, "92", "78")
);

在學會使用 HashMap 以前,我可能會作一次雙重循環:spa

List<Report> reports = new ArrayList<>();
for (Student student : students) {
    for (Score score : scores) {
        if (!student.getId().equals(score.getStudentId())) {
            continue;
        }

        reports.add(
                new Report(student.getId(), student.getName(), score.getMathScore(), score.getEnglishScore())
        );
        break;
    }
}

時間複雜度最差的狀況下是O(n * m)code

可是使用 HashMap 來改善程序,就能獲得不錯的效果:blog

Map<Long, Student> map = new HashMap<>();
for (Student student : students) {
    map.put(student.getId(), student);
}

List<Report> reports = new ArrayList<>();
for (Score score : scores) {
    Student student = map.get(score.getStudentId());
    if(student == null){ // 避免 NPE
        continue;
    }
    reports.add(
            new Report(student.getId(), student.getName(), score.getMathScore(), score.getEnglishScore())
    );
}

雙重循環,變成了兩次循環,時間複雜度是O(n + m)

顯然要比前面的方法效果要好一些。筆者寫了測試代碼分別測試兩個方法的效率,在 10w 數據下,執行時間以下:

clipboard.png

差距好像挺大。想了解爲何 HashMap 可以獲得如此好的效果,能夠看個人這篇文章:HashMap 淺析 —— LeetCode Two Sum 刷題總結。若是讀者有更好的解法歡迎留言交流,筆者水平有限,在算法上研究很少。

10w 數據的測試源碼見下方,各位讀者能夠自行試驗下效果:

package com.xiangyu.demo.hashmap;

import com.xiangyu.java.hashmap.Report;
import com.xiangyu.java.hashmap.Score;
import com.xiangyu.java.hashmap.Student;
import org.junit.Before;
import org.junit.Test;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class HashMapTest {
    private List<Student> students = new ArrayList<>();

    private List<Score> scores = new ArrayList<>();

    @Before
    public void before() {
        // 每一個list 裏放 10w 數據
        for (long i = 0; i < 100000; i++) {
            students.add(new Student(i, "test"));
            scores.add(new Score(i, "95", "95"));
        }
    }

    @Test
    public void TestHashMap() {
        Map<Long, Student> map = new HashMap<>();
        for (Student student : students) {
            map.put(student.getId(), student);
        }

        List<Report> reports = new ArrayList<>();
        for (Score score : scores) {
            Student student = map.get(score.getStudentId());
            if (student == null) {
                continue;
            }
            reports.add(
                    new Report(student.getId(), student.getName(), score.getMathScore(), score.getEnglishScore())
            );
        }
        System.out.println(reports.size());
    }

    @Test
    public void testFor2() {
        List<Report> reports = new ArrayList<>();
        for (Student student : students) {
            for (Score score : scores) {
                if (!student.getId().equals(score.getStudentId())) {
                    continue;
                }

                reports.add(
                        new Report(student.getId(), student.getName(), score.getMathScore(), score.getEnglishScore())
                );
                break;
            }
        }
        System.out.println(reports.size());
    }
}
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