如何创建嵌套TreeMaps？

这与我之前的问题有些相关，但我意识到我需要在问题的早期处理嵌套问题，所以这里是我的位置。（我也不是一个初学Java的人，所以请耐心等待）。如何创建嵌套TreeMaps？

我正在为5个房间创建一个简单的预订系统，这个系统将采用姓名，时间，房间号和日期。这必须使用嵌套的TreeMaps来完成。下面是我看到的数据，其中paretheses代表映像树的边界布局：

（日，（房间＃，（时间，名称）））

据我所看到的，我需要一个TreeMap用于时间和名称，每个房间一个，然后每天一个。这意味着每个房间每天一个时间/名称树形图，这意味着1 x 5 x 7 = 35个TreeMaps。就像这样：

{Mon, [Room 1, (0600, NameA 
        0630, NameB 
        0700, NameC) 
     Room 2, (0600, NameD 
        0630, NameE) 
     Room 3, (0600, NameF 
        0630, NameG)] 
Tues, [Room 1, (0600, Name1 
        0630, Name2) 
     Room 2, (0600, Name3 
        0630, Name4 
        0700, Name5)]}

（不同类型的支架表示嵌套的树状图的边界）

已经得出这个结论，我的下一个问题是通过迭代循环创建所有这些树状图。我似乎无法使用for循环动态生成TreeMaps，因为我无法将计数器的变量编号粘贴到新创建的TreeMap名称上。

我也有这样的：

TreeMap keyDay = new TreeMap(); 
TreeMap keyRoom = new TreeMap(); 
TreeMap keyTime = new TreeMap();

但它只有三个，这显然是不够的，允许键重复 - 例如对于任何新条目'0900'（时间键）或例如'房间1'（房间钥匙）将覆盖旧房间。

有没有人有任何建议？将不胜感激:)

来源

2011-03-27 melat0nin

为什么在这个世界上你非常喜欢TreeMaps？您需要重新设计您的应用程序，直到它不晚。三重嵌套的TreeMap就是坏的。真的，真的很糟糕。我相信你可以找到更好的解决方案。 – xappymah 2011-03-27 18:25:06

你想为此使用一个数据库。 – 2011-03-27 18:37:07

都不是一个选项，恐怕 – melat0nin 2011-03-27 19:06:21

我同意它真的超级巨大的坏。但是如果你想实现这个功能，我推荐使用HashMaps。这里是实现你想要的功能的类：

public class Reservations { 

static final int DAY_SUN = 0, DAY_MON = 1, DAY_TUE = 2, DAY_WED = 3, DAY_THU = 4, DAY_FRI = 5, DAY_SAT = 6; 
static final int ROOM_1 = 0, ROOM_2 = 1, ROOM_3 = 2, ROOM_4 = 3, ROOM_5 = 4; 
private static HashMap<Integer[], String[]> hMap = new HashMap<Integer[], String[]>(); 


static String [] getStringForValue(Integer[] i){ 
    return hMap.get(i); 
} 

static TreeSet<String> getOrderedOutputStrings(){ 
    TreeSet<String> set = new TreeSet<String>(); 
    for(Entry<Integer[],String[]> e : hMap.entrySet()){ 
     int day_int = Reservations.getDay(e.getKey()); 
     int room_int = Reservations.getRoom(e.getKey()); 
     int time = Reservations.getTime(e.getValue()); 
     String name = Reservations.getGuestName(e.getValue()); 
     String day = Reservations.dayToString(day_int); 
     String room = Reservations.roomToString(room_int); 
     if(time > 0) 
     set.add("DAY: " + "(" + day_int + ")" + day + " (" + room_int + ")"+ "ROOM: " + room + " :: " + name + " @ " + time); 
    } 
    return set; 
} 



static void setupMap() { 
    for (int day = 0; day < 7; day++) { 
     for (int room = 0; room < 5; room++) { 
      addGuest(day, room, (int)(Math.random()*1000), "Bob TestCase"); 
     } 
    } 
} 

static void addGuest(int day, int room, int time, String name) { 
    Integer[] ref = new Integer[2]; 
    ref[0] = day; 
    ref[1] = room; 
    String[] s = new String[2]; 
    s[0] = Integer.toString(time); 
    s[1] = name; 
    hMap.put(ref, s); 
} 

static String[] lookupRoom(int day, int room) { 
    Integer[] i = new Integer[2]; 
    i[0] = day; 
    i[1] = room; 
    return hMap.get(i); 
} 

static int getDay(Integer[] i){ 
    return i[0]; 
} 

static int getRoom(Integer[] i){ 
    return i[1]; 
} 

static int getTime(String[] s) { 
    return Integer.parseInt(s[0]); 
} 

public static String getGuestName(String[] s) { 
    return s[1]; 
} 

public static String dayToString(int i){ 
    switch(i){ 
     case 0: 
      return "SUNDAY"; 
     case 1: 
      return "MONDAY"; 
     case 2: 
      return "TUESDAY"; 
     case 3: 
      return "WEDNESDAY"; 
     case 4: 
      return "THURSDAY"; 
     case 5: 
      return "FRIDAY"; 
     case 6: 
      return "SATURDAY"; 
     default: 
      return null; 
    } 
} 

public static String roomToString(int i){ 
    switch(i){ 
     case 0: 
      return "ROOM ONE"; 
     case 1: 
      return "ROOM TWO"; 
     case 2: 
      return "ROOM THREE"; 
     case 3: 
      return "ROOM FOUR"; 
     case 4: 
      return "ROOM FIVE"; 
     default: 
      return null; 
    } 
}

}

这里是一个主要运行的预订类：

public class Main { 

/** 
* @param args the command line arguments 
*/ 
public static void main(String[] args) { 
    Reservations.setupMap(); //Run to test TODO: remove setupMap() from Reservation class 
    Reservations.addGuest(Reservations.DAY_MON, Reservations.ROOM_2, 1230, "John Doe"); 

    TreeSet<String> set = new TreeSet<String>(); 
    for(String s: Reservations.getOrderedOutputStrings()){ 
     System.out.println(s + "\n"); 
    } 


} 
}

，最后使用SetupMap和单一入口，它产生：

DAY: (0)SUNDAY (0)ROOM: ROOM ONE :: Bob TestCase @ 423 

DAY: (0)SUNDAY (1)ROOM: ROOM TWO :: Bob TestCase @ 637 

DAY: (0)SUNDAY (2)ROOM: ROOM THREE :: Bob TestCase @ 731 

DAY: (0)SUNDAY (3)ROOM: ROOM FOUR :: Bob TestCase @ 424 

DAY: (0)SUNDAY (4)ROOM: ROOM FIVE :: Bob TestCase @ 427 

DAY: (1)MONDAY (0)ROOM: ROOM ONE :: Bob TestCase @ 335 

DAY: (1)MONDAY (1)ROOM: ROOM TWO :: Bob TestCase @ 580 

DAY: (1)MONDAY (1)ROOM: ROOM TWO :: John Doe @ 1230 

DAY: (1)MONDAY (2)ROOM: ROOM THREE :: Bob TestCase @ 700 

DAY: (1)MONDAY (3)ROOM: ROOM FOUR :: Bob TestCase @ 556 

DAY: (1)MONDAY (4)ROOM: ROOM FIVE :: Bob TestCase @ 92 

DAY: (2)TUESDAY (0)ROOM: ROOM ONE :: Bob TestCase @ 84 

DAY: (2)TUESDAY (1)ROOM: ROOM TWO :: Bob TestCase @ 964 

DAY: (2)TUESDAY (2)ROOM: ROOM THREE :: Bob TestCase @ 981 

DAY: (2)TUESDAY (3)ROOM: ROOM FOUR :: Bob TestCase @ 294 

DAY: (2)TUESDAY (4)ROOM: ROOM FIVE :: Bob TestCase @ 22 

DAY: (3)WEDNESDAY (0)ROOM: ROOM ONE :: Bob TestCase @ 216 

DAY: (3)WEDNESDAY (1)ROOM: ROOM TWO :: Bob TestCase @ 838 

DAY: (3)WEDNESDAY (2)ROOM: ROOM THREE :: Bob TestCase @ 198 

DAY: (3)WEDNESDAY (3)ROOM: ROOM FOUR :: Bob TestCase @ 494 

DAY: (3)WEDNESDAY (4)ROOM: ROOM FIVE :: Bob TestCase @ 796 

DAY: (4)THURSDAY (0)ROOM: ROOM ONE :: Bob TestCase @ 617 

DAY: (4)THURSDAY (1)ROOM: ROOM TWO :: Bob TestCase @ 866 

DAY: (4)THURSDAY (2)ROOM: ROOM THREE :: Bob TestCase @ 799 

DAY: (4)THURSDAY (3)ROOM: ROOM FOUR :: Bob TestCase @ 242 

DAY: (4)THURSDAY (4)ROOM: ROOM FIVE :: Bob TestCase @ 178 

DAY: (5)FRIDAY (0)ROOM: ROOM ONE :: Bob TestCase @ 491 

DAY: (5)FRIDAY (1)ROOM: ROOM TWO :: Bob TestCase @ 16 

DAY: (5)FRIDAY (2)ROOM: ROOM THREE :: Bob TestCase @ 838 

DAY: (5)FRIDAY (3)ROOM: ROOM FOUR :: Bob TestCase @ 63 

DAY: (5)FRIDAY (4)ROOM: ROOM FIVE :: Bob TestCase @ 860 

DAY: (6)SATURDAY (0)ROOM: ROOM ONE :: Bob TestCase @ 73 

DAY: (6)SATURDAY (1)ROOM: ROOM TWO :: Bob TestCase @ 759 

DAY: (6)SATURDAY (2)ROOM: ROOM THREE :: Bob TestCase @ 15 

DAY: (6)SATURDAY (3)ROOM: ROOM FOUR :: Bob TestCase @ 115 

DAY: (6)SATURDAY (4)ROOM: ROOM FIVE :: Bob TestCase @ 21

该结果是在不到一秒钟内生成的。我保证这比嵌套的TreeMaps更高效。祝你好运！

来源

2011-03-27 20:32:54 Josh

对于问题的分析，使用Tree或Hashmap是完全不相关的。问题在于，Map是否合适，如果是，则Map必须嵌套在一起。 – Ingo 2011-03-27 20:42:08

虽然我同意你从来没有真正回答过这个问题，但我正在给你+1努力并加倍努力。 – bconneen 2011-03-27 21:49:01

我同意bconneen - 我不能使用你的代码，但是感谢你抽出时间写下它！ – melat0nin 2011-03-28 17:18:07

人们会认为这一天实际上是一个枚举，所以顶级的地图应该是一个EnumMap的

Map<Day, Map<String, Map<Integer, String>>> bookings = new EnumMap<Day, Map<String, Map<Integer, String>>>(Day.class);

来源

2011-03-27 18:45:17 MeBigFatGuy

不是一个真正的答案，而是一种思想;）

的确，正如xappymah在评论中指出的那样，更自然的方法是为您的任务发明特定于领域的类型。像：

interface Room { 
    String getName(); 
    public void bookRoom(Booking booking) throws AlreadyBookedException; 
} 

interface Person { 
    String getName(); 
} 

class Interval { 
    Date startTime; 
    long duration; 
} 

interface Booking { 
    Interval getInterval(); 
    Room getBookedRoom() throws NotBookedYetException; 
    Set<Person> getParticipants(); 
} 

interface BookingUtils { 
    Set<Booking> getBookingsForInterval(Interval interval); 
    Set<Booking> getBookingsOfRoom(Room room); 
}

来源

2011-03-27 18:50:03

唉，我必须使用TreeMaps;这是针对大学的练习：/ – melat0nin 2011-03-27 19:05:51

如何创建嵌套TreeMaps？

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