如何使用比较器接口按地图的字符串键对地图进行排序

Question

我知道以前也曾问过类似的问题，但是我有一个特定的问题，我不能误解。

我想按星期几对地图进行排序，其中星期几是地图的关键。 您如何在compare方法中比较没有自然顺序的字符串？ 排序末尾的地图应该是：星期一，星期二，星期三，星期四。

到目前为止，这是我的代码，但是我只能使用compare方法。

    class OpeningTimes {

       private String openingTime;
       private String closingTime;
       public String getOpeningTime() {
           return openingTime;
       }
       public void setOpeningTime(String openingTime) {
           this.openingTime = openingTime;
       }
       public String getClosingTime() {
           return closingTime;
       }
       public void setClosingTime(String closingTime) {
           this.closingTime = closingTime;
       }

       @Override
       public String toString() {
           return new StringBuilder( "Opening Times: " + this.openingTime + " Closing Times: " + this.closingTime).toString();
       }
   }

比较：

class OpeningTimesComparator implements Comparator<Map.Entry<String, OpeningTimes>>{

   @Override
   public int compare(Entry<String, OpeningTimes> o1, Entry<String, OpeningTimes> o2) {
       return 0; // what is the logic?
   }}

亚军：

public class TestClass {

    public static void main(String[] args) {

        Map<String, OpeningTimes> openingTimesMap = new TreeMap<String, OpeningTimes>();

        OpeningTimes openTime1 = new OpeningTimes();
        openTime1.setOpeningTime("9PM");
        openTime1.setClosingTime("10PM");

        OpeningTimes openTime2 = new OpeningTimes();
        openTime2.setOpeningTime("11PM");
        openTime2.setClosingTime("9PM");

        OpeningTimes openTime3 = new OpeningTimes();
        openTime3.setOpeningTime("13PM");
        openTime3.setClosingTime("14PM");

        OpeningTimes openTime4 = new OpeningTimes();
        openTime4.setOpeningTime("15PM");
        openTime4.setClosingTime("13PM");

        openingTimesMap.put("Tue", openTime2);
        openingTimesMap.put("Thu", openTime4);
        openingTimesMap.put("Mon", openTime1);
        openingTimesMap.put("Wed", openTime3);


        for (Entry<String, OpeningTimes> openingTimesSingle : openingTimesMap.entrySet()) {
            System.out.println("Key: " + openingTimesSingle.getKey() + " Value: " + openingTimesMap.get(openingTimesSingle.getKey()));
        }

        List<Map.Entry<String, OpeningTimes>> list = new ArrayList<Map.Entry<String, OpeningTimes>>(openingTimesMap.entrySet());

        Collections.sort(list, new OpeningTimesComparator());

        for (Entry<String, OpeningTimes> openingTimesSingle : openingTimesMap.entrySet()) {
            System.out.println("Key: " + openingTimesSingle.getKey() + " Value: " + openingTimesMap.get(openingTimesSingle.getKey()));
        }
    }
}

提前致谢

Answer 1

如果您不必使用其他一些包含自定义订单值的静态地图，则枚举将是一个很好的解决方案。

定义一个这样的枚举

enum Week{
  SUN,MON,TUE,WED,THU,FRI,SAT   
}

创建一个EnumMap实例并添加条目，并在keySet进行迭代时，将按枚举中提到的顺序检索条目

    Map<Week, String> map = new EnumMap<Week, String>(Week.class);

        map.put(Week.WED, "Wednesday");
        map.put(Week.SUN, "Sunday");
        map.put(Week.MON, "Monday");
        map.put(Week.THU, "Thursday");
        map.put(Week.SAT, "Saturday");
        map.put(Week.FRI, "Friday");
        map.put(Week.TUE, "Tuesday");


        for(Week week : map.keySet()){
            System.out.println(week + " -- " +map.get(week));
        }

输出：

SUN -- Sunday
MON -- Monday
TUE -- Tuesday
WED -- Wednesday
THU -- Thursday
FRI -- Friday
SAT -- Saturday

我在枚举中定义的顺序相同。 当然，仅当Map中的键范围有限时（例如，一周中的几天）， EnumMap才有用且高效。 希望这可以帮助 ：）

选项2：-

这是在假设inputKeys完全由您管理的前提下进行的。 请注意这一点，因为如果给定密钥不存在映射，它可能会给您NPE。

private static Map<String,Integer> orderMappings = new HashMap<>();

    static{
        orderMappings.put("Sunday", 1);
        orderMappings.put("Monday", 2);
        orderMappings.put("Tuesday", 3);
        orderMappings.put("Wednesday", 4);
        orderMappings.put("Thursday", 5);
        orderMappings.put("Friday", 6);
        orderMappings.put("Saturday", 7);
    }

    class OrderComparator implements Comparator<String>{

        @Override
        public int compare(String key1, String key2) {
            int p1 = orderMappings.get(key1);
            int p2 =  orderMappings.get(key2);
            return p1-p2;
        }

    }