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如何遍歷多個列表並使用Java 8 Lambda表達式合並兩個列表中的唯一值來創建另一個列表

[英]How to loop through multiple Lists and create another list with merging of unique values from both the lists using Java 8 lambda expressions

我正在編寫代碼以遍歷多個列表,並使用Java 8 lambda表達式合並兩個列表中的唯一值來創建另一個列表。

型號類別:

class ServiceMap{
    Integer serviceMapId;
    Integer seviceId;
}

代碼邏輯:

List<ServiceMap> listA = getServiceMaps();//Will get from database
List<Integer> listB = Arrays.asList(1, 10, 9);//Will get from client
List<ServiceMap> listC = new ArrayList<>();//Building it merging of both lists above

listA.stream().forEach(e -> {
    if (listB.parallelStream().noneMatch(x -> x == e.getServiceId())) {
        listC.add(new ServiceMap(e.getServiceId()));
        return;
    }

    listB.stream().forEach(x -> {
        if (listC.stream().anyMatch(e2->e2.getServiceId() == x)) {
            return;
        }
        if (x == e.getServiceId()) {
            listC.add(new ServiceMap(e.getServiceId()));
        } else {
            listC.add(new ServiceMap(x));
        }
    });

});
listC.stream().forEach(x -> System.out.println(x));

使用Java Lambda表達式編寫代碼是否有效?

您可以流式傳輸每個列表,對其distinct進行應用並收集:

List<Integer> result = 
    Stream.concat(listA.stream(), listB.stream())
          .distinct()
          .collect(Collectors.toList());

您也應該這樣使用;

Stream<Integer> streamOfServiceMapIds = listA.stream().map(ServiceMap::getSeviceId);
List<ServiceMap> collectedList = Stream.concat(streamOfServiceMapIds, listB.stream())
        .distinct()
        .map(ServiceMap::new)
        .collect(Collectors.toList());

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