public class WordFrequency {
public static String sentence = "one three two two three three four four four";
public static Map<String,Integer> map;
public static void main(String[] args) {
map = new HashMap<>();
String[] words = sentence.split("\\s");
for (String word : words) {
Integer count = map.get(word);
if (count == null) {
count = 1;
} else {
++count;
}
map.put(word,count);
}
Comparator<String> myComparator = new Comparator<String>() {
@Override
public int compare(String s1,String s2) {
if (map.get(s1) < map.get(s2)) {
return -1;
} else if (map.get(s1) > map.get(s2)) {
return 1;
} else {
return 0;
}
}
};
SortedMap<String,Integer> sortedMap = new TreeMap<String,Integer>(myComparator);
System.out.println("Before sorting: " + map);
sortedMap.putAll(map);
System.out.println("After Sorting based on value:" + sortedMap);
}
}
输出：
Before sorting: {two=2,one=1,three=3,four=3}
After sorting based on value:{one=1,two=2,three=3}
预期产出：
{one=1,four=3,three=3}
解决方法
Note that the ordering maintained by a tree map,like any sorted map,and whether or not an explicit comparator is provided,must be consistent with equals if this sorted map is to correctly implement the Map interface. (See Comparable or Comparator for a precise definition of consistent with equals.) This is so because the Map interface is defined in terms of the equals operation,but a sorted map performs all key comparisons using its compareTo (or compare) method,so two keys that are deemed equal by this method are,from the standpoint of the sorted map,equal. The behavior of a sorted map is well-defined even if its ordering is inconsistent with equals; it just fails to obey the general contract of the Map interface.
您可以通过在值相等时比较键来解决此问题：
Comparator<String> myComparator = new Comparator<String>() {
@Override
public int compare(String s1,String s2) {
if (map.get(s1) < map.get(s2)) {
return -1;
} else if (map.get(s1) > map.get(s2)) {
return 1;
} else {
return s1.compareTo(s2);
}
}
};
这应该给你一个输出：
After sorting based on value:{one=1,three=3}
由于四个<三个基于字符串的自然顺序.
