HashMap 常见应用：实现 SQL JOIN

在我的上一篇文章中，讲到了我自己初步认识 HashMap 的一个经验分享：HashMap 浅析 —— LeetCode Two Sum 刷题总结。作为一个 CRUD 工程师，平时很少接触到基础组件的涉及，那么是不是很难有机会用到 HashMap 呢？今天，就举一个常见的查询例子，来看看我们如何使用 HashMap 来提高代码的效率。已知一个 Student 类：public class Student { private Long id; private String name; public Student(Long id, String name) { this.id = id; this.name = name; } // —Getters And Setters—}和一个 Score 类：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—}我们需要把 Student 和 Score 合并到一起，即类 Report：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; } }看类的属性我们就明白了，这里其实相当于在 Student 和 Score 之间做一个 Join，得到 Report。这是我们在编程中常见的场景（例如查询了订单中心，用户中心，支付中心，合并各个中心返回的结果形成一个表单，因为各个中心是独立的微服务，无法使用 SQL JOIN）。现有两个 List：List<Student> 和 List<Score>：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”));在学会使用 HashMap 之前，我可能会做一次双重循环：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; }}时间复杂度最差的情况下是O(n * m)。但是使用 HashMap 来改善程序，就能得到不错的效果：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()) );}双重循环，变成了两次循环，时间复杂度是O(n + m)。显然要比前面的方法效果要好一些。笔者写了测试代码分别测试两个方法的效率，在 10w 数据下，执行时间如下：差距好像挺大。想了解为什么 HashMap 能够得到如此好的效果，可以看我的这篇文章：HashMap 浅析 —— LeetCode Two Sum 刷题总结。如果读者有更好的解法欢迎留言交流，笔者水平有限，在算法上研究不多。10w 数据的测试源码见下方，各位读者可以自行试验下效果：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() { // 每个list 里放 10w 数据 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()); }}