LinkedHashMap排序

Question

我想基於它的值排序LinkedHashMap。我不明白的是結果。它似乎只需要兩個鍵進行排序。任何指向我缺少什麼？

public class test { 

    public static void main(String[] args) { 
     // TODO Auto-generated method stub 


     LinkedHashMap<Integer, Integer> sorting = new LinkedHashMap<Integer, Integer>(); 
     sorting.put(1, 100); 
     sorting.put(10, 100); 
     sorting.put(20, 200); 
     sorting.put(30, 100); 
     sorting.put(40, 100); 
     sorting.put(50, 200); 

     for (Entry<Integer, Integer> entry : sorting.entrySet()) { 
      Integer key = entry.getKey(); 
      Integer value = entry.getValue(); 
      System.out.println("LINKED UNSORTED === key: "+ key + '\t' + "Value: " + value); 
       // do stuff 
      } 


     Comparator<Integer> comparator = new ValueCom(sorting); 
     TreeMap<Integer, Integer> sortedMap =new TreeMap<Integer, Integer>(comparator); 
     sortedMap.putAll(sorting);  


     for (Entry<Integer, Integer> entry : sortedMap.entrySet()) { 
      Integer key = entry.getKey(); 
      Integer value = entry.getValue(); 
      System.out.println("SORTED === key: "+ key + '\t' + "Value: " + value); 
       // do stuff 
      } 

    } 

} 


class ValueCom implements Comparator<Integer> { 
    LinkedHashMap<Integer, Integer> map = new LinkedHashMap<Integer, Integer>(); 

public ValueCom(HashMap<Integer, Integer> map) { 
     this.map.putAll(map); 
    } 

public int compare(Integer keyA, Integer keyB){ 

    return map.get(keyB).compareTo(map.get(keyA)); 
} 

} 

輸出的電流如上

LINKED UNSORTED === key: 1 Value: 100 
LINKED UNSORTED === key: 10 Value: 100 
LINKED UNSORTED === key: 20 Value: 200 
LINKED UNSORTED === key: 30 Value: 100 
LINKED UNSORTED === key: 40 Value: 100 
LINKED UNSORTED === key: 50 Value: 200 
SORTED === key: 20 Value: 200 
SORTED === key: 1 Value: 100 

Answer 1

解釋在TreeMap密鑰的唯一性是由傳遞Comparator確定。由於您的Comparator比較了原始Map的值，所以具有相同值的所有條目都被TreeMap視爲具有相同的密鑰，因此對於每個唯一值，只有一個條目被添加到TreeMap。

如果要通過值排序的Map，您可以使用您的排序Map一個LinkedHashMap，並確保你將這些條目中Map所需的順序。

Answer 2

問題是你正在對值的基礎上進行排序，如果兩個值相等，那麼TreeMap將只爲該鍵創建單個條目，通過查看下面的代碼可以更好地理解。下面的代碼將解決這個問題，讓我知道，如果有什麼失敗：

修改compare()方法，如下面ValueCom類：

public int compare(Integer i1, Integer i2) { 
     if(map.get(i1).equals(map.get(i2))){ 
      return 1; //To make sure, If values are equal then still there should be entry in TreeMap 
     } 
     return map.get(i1).compareTo(map.get(i2)); 
    } 

現在，值相等的情況下，也處理。 :)

Answer 3

下面的代碼給你你想要什麼

public class test { 

public static void main(String[] args) { 
    // TODO Auto-generated method stub 


    LinkedHashMap<Integer, Integer> sorting = new LinkedHashMap<Integer, Integer>(); 
    sorting.put(10, 100); 
    sorting.put(1, 100); 
    sorting.put(20, 200); 
    sorting.put(40, 100); 
    sorting.put(30, 100); 
    sorting.put(50, 200); 

    for (Entry<Integer, Integer> entry : sorting.entrySet()) { 
     Integer key = entry.getKey(); 
     Integer value = entry.getValue(); 
     System.out.println("LINKED UNSORTED === key: "+ key + '\t' + "Value: " + value); 
      // do stuff 
     } 


    Comparator<Integer> comparator = new ValueCom(); 
    TreeMap<Integer, Integer> sortedMap =new TreeMap<Integer, Integer>(comparator); 
    sortedMap.putAll(sorting);  


    for (Entry<Integer, Integer> entry : sortedMap.entrySet()) { 
     Integer key = entry.getKey(); 
     Integer value = entry.getValue(); 
     System.out.println("SORTED === key: "+ key + '\t' + "Value: " + value); 
      // do stuff 
     } 

} 

} 


class ValueCom implements Comparator<Integer> { 
public int compare(Integer keyA, Integer keyB){ 

return keyA.compareTo(keyB); 
} 

}