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你如何将每个收藏的物品与另一个收藏的物品相关联?

permutation collections java

Alberto Zaccagni · 技术社区 · 17 年前

Collection<Room> roomsFromA = getRoomsFromA();
Collection<Room> roomsFromB = getRoomsFromB();
for(Room roomA : roomsFromA){
    for(Room roomB : roomsFromB){
        //add roomA and roomB to something, this is not important for what I need
        //the important part is how you handle the part before
        //especially if Collection objects number grows beyond two
    }
}

编辑2:我会尽量解释得更好,很抱歉问题不清楚。用户要求一间双人房和一间单人房。酒店有3间双人房和4间单人房可供选择。

8 回复 | 直到 17 年前

willcodejavaforfood 17 年前

this example

AgileJon 17 年前

这些收藏品排得整整齐齐吗?

HashMap map = new HashMap();
for (int i=0; i<c1.Size(); i++) {
   map.put(c1[i], c2[i]);
}

José Leal 17 年前

private static class Pair<K, T> {
    private K one;
    private T two;

    public Pair(K one, T two) {
        this.one = one;
        this.two = two;
    }

    /**
     * @return the one
     */
    public K getOne() {
        return one;
    }

    /**
     * @return the two
     */
    public T getTwo() {
        return two;
    }
}

并与他们一起创建一个列表。

Steve B. 17 年前

class Room {
   public Collection<Room> getRoomsFromB { ... 
}

//收藏室

for (Room a: rooms)
{ 
   for(Room b a.getRoomsFromB){ ...

Alex Feinman 17 年前

bruno conde 17 年前

中的每个元素 collection

对两个集合进行排序(在同一个集合中
用一个循环迭代两个集合,构建一个关系对象并将其添加到一个新的集合中。

public static class Room {
    private int number;
    private String name;

    public Room(int number, String name) {
        super();
        this.number = number;
        this.name = name;
    }

    public int getNumber() {
        return number;
    }

    public String getName() {
        return name;
    }
}

public static class RoomRelation {
    private Room a;
    private Room b;

    public RoomRelation(Room a, Room b) {
        super();
        this.a = a;
        this.b = b;
    }

    public Room getA() {
        return a;
    }

    public Room getB() {
        return b;
    }

    @Override
    public String toString() {
        return a.getName() + "(" + a.getNumber() + ") " + b.getName() + "(" + b.getNumber() + ")";
    }
}

public static void main(String[] args) {

    List<Room> roomsFromA = new ArrayList<Room>();
    List<Room> roomsFromB = new ArrayList<Room>();

    roomsFromA.add(new Room(1,"Room A"));
    roomsFromA.add(new Room(2,"Room A"));

    roomsFromB.add(new Room(1,"Room B"));
    roomsFromB.add(new Room(2,"Room B"));

    Comparator<Room> c = new Comparator<Room>() {
        @Override
        public int compare(Room o1, Room o2) {
            return o1.getNumber() - o2.getNumber();
        } };

    Collections.sort(roomsFromA, c);
    Collections.sort(roomsFromB, c);

    List<RoomRelation> relations = new ArrayList<RoomRelation>();

    for (int i = 0; i < roomsFromA.size(); i++) {
        relations.add(new RoomRelation(roomsFromA.get(i), roomsFromB.get(i)));
    }

    for (RoomRelation roomRelation : relations) {
        System.out.println(roomRelation);
    }
}

ewh105 17 年前

Collection<Room> roomsFromA = getRoomsFromA();
Collection<Room> roomsFromB = getRoomsFromB();

Room[][] combinations = new Room[roomsFromA .size()][roomsFromB .size()];

int a = 0;
int b = 0;

for(Room roomA : roomsFromA){

   for(Room roomB : roomsFromB){
      combinations [a][b] = [roomA][roomB]; //Build up array
      b++; 
   }
   a++;
}

return combinations;

-1

Community Mohan Dere 9 年前

Cartesian product this question .