Question

我有一个以Array为值的地图。现在，所有这些数组都需要转换为列表。最后，所有列表都应合并为一个列表。

现在可用，

Map<String, Class1[]> master_map= new HashMap<>();

Class1 c1 = new Class("Name1", "Address1");
Class1 c2 = new Class("Name2", "Address2");
ClassArray[0] = c1;
ClassArray[1]= c2;
master_map.put("FirstValue", ClassArray);

Class1 c3 = new Class("Name3", "Address3");
Class1 c4 = new Class("Name4", "Address4");
ClassArray[0] = c3;
ClassArray[1]= c4;
master_map.put("SecondValue", ClassArray);}

现在我希望得到如下结果。

List<Class1> result  = new ArrayList<>();
//This above list to be as {c1, c2, c3, c4}

期待java8中的代码

Answer 1

您需要在条目流中展平内部数组：

template<typename T>
class ThreadSafeQueue
{
public:
    ThreadSafeQueue() {}
    ~ThreadSafeQueue() {}

    void Push(T d)
    {
        std::unique_lock<std::mutex> locker(mutex);

        data.push(std::move(d));

        locker.unlock();

        cond.notify_one();
    }

    T* get()
    {
        std::unique_lock<std::mutex> locker(mutex);

        cond.wait(locker, [=]() { return stop || !data.empty(); });

        if (stop && data.empty()) {
            nullptr;
        }

        T res = std::move(data.front());
        data.pop();

        return &res;
    }

    bool Empty()
    {
        std::lock_guard<std::mutex> guard(mutex);
        return data.empty();
    }

    std::queue<T> data;

    std::mutex mutex;
    std::condition_variable cond;
    bool stop = false;
};

ThreadSafeQueue<SDL_Surface*> queue;
void LoadingThread(const std::string& path) 
{
    SDL_Surface* surf = std::async(std::launch::async, IMG_Load, path.c_str()).get();
    queue.Push(surf);
}

void Texture::LoadFromFile(const std::string& path)
{
    LoadingThread(path);
    SDL_Surface* surface = nullptr;

    if (!queue.data.empty() && path != "") {
        surface = *queue.get();
        if(surface) {
            w = surface->w;
            h = surface->h;
            pixels = surface->pixels;
        } else return;
    } else return;

    glGenTextures(1, &texture);
    glBindTexture(GL_TEXTURE_2D, texture);

    glTexImage2D(GL_TEXTURE_2D, 0, GL_RGB, w, h, 0, GL_RGB, GL_UNSIGNED_BYTE, pixels);

    glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_REPEAT);
    glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_T, GL_REPEAT);
    glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER, GL_LINEAR);
    glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_LINEAR);

    glGenerateMipmap(GL_TEXTURE_2D);

    glBindTexture(GL_TEXTURE_2D, 0);
}


void Draw() 
{
    drawThread = std::make_unique<std::thread>([&]() {
        drawContext->MakeCurrent();
        renderer2D.Create(window);

        std::unique_ptr<Texture> texture = Texture::LoadPNG("Textures/test.png");

        while (!quit) {
            //glClearColor - I prefere 0 - 255 format
            Renderer::ClearColor(sinf(SDL_GetTicks() / 500.0f) * 255.0f, 0.0f, 255.0f, 255.0f);
            //glClear() color buffer bit default
            Renderer::ClearBuffers();

            //simple batch renderer 
            renderer2D.RenderClear();
            renderer2D->Draw(texture, { 0.0f, 0.0f, 50.0f, 50.0f });
            renderer2D.RenderPresent();

            window.SwapWindows();
        }
    });
}

Answer 2

您可以使用条目集遍历地图，对于每个值（即数组），可以使用Arrays.asList将其转换为列表，并将其添加到最终列表中。

    Map<String , Integer[]> hm = new HashMap<String,Integer[]>();
    List<Integer> finalList = new ArrayList<Integer>();
    for(Entry<String, Integer[]> entry:hm.entrySet()) {
        finalList.addAll(Arrays.asList(entry.getValue()));
    }

使用并行流可以像这样

    Map<String , Integer[]> hm = new HashMap<String,Integer[]>();

    List<Integer> finalList = hm.values().parallelStream().flatMap(Arrays::stream).collect(Collectors.toList())

这是整数数组的简单示例。您可以参考并得出您的解决方案。

Answer 3

另一个Java 8简单变体：

List<Class1> result  = new ArrayList<>();
master_map.entrySet().stream().forEach(element -> {
  result.addAll(Arrays.asList(element.getValue()));
    });

Answer 4

另一个例子

    List<Class1> x = new ArrayList<>();
    master_map.forEach((k,v) -> x.addAll(Arrays.asList(v)));

映射值在数组中。所有值都需要转换为列表，并将所有结果组合为结果列表

4 个答案: