Java 8 中 Map

Java 8 最大的特性无异于更多地面向函数,引入了 lambda可以更好地进行函数式编程
merge是什么？
merge() 可以这么理解：它将新的值赋值到 key （如果不存在）或更新给定的key 值对应的 value

merge源码参考

default V merge(K key, V value, BiFunction<? super V, ? super V, ? extends V> remappingFunction) {
    Objects.requireNonNull(remappingFunction);
    Objects.requireNonNull(value);
    V oldValue = this.get(key);
    V newValue = oldValue == null ? value : remappingFunction.apply(oldValue, value);
    if (newValue == null) {
        this.remove(key);
    } else {
        this.put(key, newValue);
    }

    return newValue;
}
//该方法接收三个参数，一个 key 值，一个 value，
//一个 remappingFunction ，如果给定的key不存在，它就变成了 put(key, value) 。
//但是，如果 key 已经存在一些值，
//我们  remappingFunction 可以选择合并的方式，
//然后将合并得到的 newValue 赋值给原先的 key。

merge() 怎么用？
假设有一个学生成绩对象的列表，对象包含学生姓名、科目、科目分数三个属性，要求求得每个学生的总成绩。加入列表如下：

private List<StudentScore> buildATestList() {
    List<StudentScore> studentScoreList = new ArrayList<>();
    StudentScore studentScore1 = new StudentScore() {{
        setStuName("张三");
        setSubject("语文");
        setScore(70);
    }};
    StudentScore studentScore2 = new StudentScore() {{
        setStuName("张三");
        setSubject("数学");
        setScore(80);
    }};
    StudentScore studentScore3 = new StudentScore() {{
        setStuName("张三");
        setSubject("英语");
        setScore(65);
    }};
    StudentScore studentScore4 = new StudentScore() {{
        setStuName("李四");
        setSubject("语文");
        setScore(68);
    }};
    StudentScore studentScore5 = new StudentScore() {{
        setStuName("李四");
        setSubject("数学");
        setScore(70);
    }};
    StudentScore studentScore6 = new StudentScore() {{
        setStuName("李四");
        setSubject("英语");
        setScore(90);
    }};
    StudentScore studentScore7 = new StudentScore() {{
        setStuName("王五");
        setSubject("语文");
        setScore(80);
    }};
    StudentScore studentScore8 = new StudentScore() {{
        setStuName("王五");
        setSubject("数学");
        setScore(85);
    }};
    StudentScore studentScore9 = new StudentScore() {{
        setStuName("王五");
        setSubject("英语");
        setScore(70);
    }};

    studentScoreList.add(studentScore1);
    studentScoreList.add(studentScore2);
    studentScoreList.add(studentScore3);
    studentScoreList.add(studentScore4);
    studentScoreList.add(studentScore5);
    studentScoreList.add(studentScore6);
    studentScoreList.add(studentScore7);
    studentScoreList.add(studentScore8);
    studentScoreList.add(studentScore9);

    return studentScoreList;
}

常规做法

ObjectMapper objectMapper = new ObjectMapper();
List<StudentScore> studentScoreList = buildATestList();

Map<String, Integer> studentScoreMap = new HashMap<>();
studentScoreList.forEach(studentScore -> {
    if (studentScoreMap.containsKey(studentScore.getStuName())) {
        studentScoreMap.put(studentScore.getStuName(), 
                            studentScoreMap.get(studentScore.getStuName()) + studentScore.getScore());
    } else {
        studentScoreMap.put(studentScore.getStuName(), studentScore.getScore());
    }
});

System.out.println(objectMapper.writeValueAsString(studentScoreMap));

结果如下

 {"李四":228,"张三":215,"王五":235}

merge() 实现

Map<String, Integer> studentScoreMap2 = new HashMap<>();
studentScoreList.forEach(studentScore -> studentScoreMap2.merge(
  studentScore.getStuName(),
  studentScore.getScore(),
  Integer::sum));

System.out.println(objectMapper.writeValueAsString(studentScoreMap2));

结果如下

{"李四":228,"张三":215,"王五":235}

使用场景

比如分组求和这类的操作，虽然 stream 中有相关 groupingBy() 方法，但如果你想在循环中做一些其他操作的时候，merge() 还是一个挺不错的选择的。