我试图根据LinkedHashMap的值对其进行排序。我不明白的是结果。排序似乎只需要两个键。关于我错过了什么有什么建议吗?
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));
}
} 当前的输出如上文所解释。
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发布于 2016-06-07 06:45:58
在TreeMap中,键的唯一性由传递的Comparator决定。因为您的Comparator比较原始Map的值,所以TreeMap认为具有相同值的所有条目都具有相同的键,因此对于每个唯一值,只向TreeMap添加一个条目。
如果您想要按值排序的Map,可以为排序的Map使用LinkedHashMap,并确保按所需的顺序将条目放在Map中。
发布于 2016-06-07 07:10:45
问题是,您正在根据值进行排序,如果两个值相等,那么TreeMap将只为该键创建一个条目,您可以通过查看下面的代码来更好地理解。下面的代码将解决这个问题,如果有任何问题发生,请告诉我:
修改compare()类中的ValueCom方法如下:
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));
}现在,同样的情况也被处理了。:)
发布于 2016-06-07 07:14:45
下面的代码给出了您想要的
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);
}
} https://stackoverflow.com/questions/37672213
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