自己怎么1做网站常用的搜索引擎有哪些?
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。