重要说明：

此代码可以以多种方式中断。如果您打算使用提供的代码，请务必阅读注释以了解其含义。例如，值不能再通过其键检索。（get始终返回null。）

这似乎比前面所有的都容易得多。按如下方式使用TreeMap：

public class Testing {
    public static void main(String[] args) {
        HashMap<String, Double> map = new HashMap<String, Double>();
        ValueComparator bvc = new ValueComparator(map);
        TreeMap<String, Double> sorted_map = new TreeMap<String, Double>(bvc);

        map.put("A", 99.5);
        map.put("B", 67.4);
        map.put("C", 67.4);
        map.put("D", 67.3);

        System.out.println("unsorted map: " + map);
        sorted_map.putAll(map);
        System.out.println("results: " + sorted_map);
    }
}

class ValueComparator implements Comparator<String> {
    Map<String, Double> base;

    public ValueComparator(Map<String, Double> base) {
        this.base = base;
    }

    // Note: this comparator imposes orderings that are inconsistent with
    // equals.
    public int compare(String a, String b) {
        if (base.get(a) >= base.get(b)) {
            return -1;
        } else {
            return 1;
        } // returning 0 would merge keys
    }
}

输出：

unsorted map: {D=67.3, A=99.5, B=67.4, C=67.4}
results: {D=67.3, B=67.4, C=67.4, A=99.5}

public class Test {
  public static void main(String[] args) {
    TreeMap<Integer, String> hm=new TreeMap();
    hm.put(3, "arun singh");
    hm.put(5, "vinay singh");
    hm.put(1, "bandagi singh");
    hm.put(6, "vikram singh");
    hm.put(2, "panipat singh");
    hm.put(28, "jakarta singh");

    ArrayList<String> al=new ArrayList(hm.values());
    Collections.sort(al, new myComparator());

    System.out.println("//sort by values \n");
    for(String obj: al){
        for(Map.Entry<Integer, String> map2:hm.entrySet()){
            if(map2.getValue().equals(obj)){
                System.out.println(map2.getKey()+" "+map2.getValue());
            }
        } 
     }
  }
}

class myComparator implements Comparator{
    @Override
    public int compare(Object o1, Object o2) {
       String o3=(String) o1;
       String o4 =(String) o2;
       return o3.compareTo(o4);
    }   
}

输出，输出=

//sort by values 

3 arun singh
1 bandagi singh
28 jakarta singh
2 panipat singh
6 vikram singh
5 vinay singh

我合并了user157196和Carter Page的解决方案：

class MapUtil {

    public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue( Map<K, V> map ){
        ValueComparator<K,V> bvc =  new ValueComparator<K,V>(map);
        TreeMap<K,V> sorted_map = new TreeMap<K,V>(bvc);
        sorted_map.putAll(map);
        return sorted_map;
    }

}

class ValueComparator<K, V extends Comparable<? super V>> implements Comparator<K> {

    Map<K, V> base;
    public ValueComparator(Map<K, V> base) {
        this.base = base;
    }

    public int compare(K a, K b) {
        int result = (base.get(a).compareTo(base.get(b)));
        if (result == 0) result=1;
        // returning 0 would merge keys
        return result;
    }
}

    static <K extends Comparable<? super K>, V extends Comparable<? super V>>
    Map sortByValueInDescendingOrder(final Map<K, V> map) {
        Map re = new TreeMap(new Comparator<K>() {
            @Override
            public int compare(K o1, K o2) {
                if (map.get(o1) == null || map.get(o2) == null) {
                    return -o1.compareTo(o2);
                }
                int result = -map.get(o1).compareTo(map.get(o2));
                if (result != 0) {
                    return result;
                }
                return -o1.compareTo(o2);
            }
        });
        re.putAll(map);
        return re;
    }
    @Test(timeout = 3000l, expected = Test.None.class)
    public void testSortByValueInDescendingOrder() {
        char[] arr = "googler".toCharArray();
        Map<Character, Integer> charToTimes = new HashMap();
        for (int i = 0; i < arr.length; i++) {
            Integer times = charToTimes.get(arr[i]);
            charToTimes.put(arr[i], times == null ? 1 : times + 1);
        }
        Map sortedByTimes = sortByValueInDescendingOrder(charToTimes);
        Assert.assertEquals(charToTimes.toString(), "{g=2, e=1, r=1, o=2, l=1}");
        Assert.assertEquals(sortedByTimes.toString(), "{o=2, g=2, r=1, l=1, e=1}");
        Assert.assertEquals(sortedByTimes.containsKey('a'), false);
        Assert.assertEquals(sortedByTimes.get('a'), null);
        Assert.assertEquals(sortedByTimes.get('g'), 2);
        Assert.assertEquals(sortedByTimes.equals(charToTimes), true);
    }

最干净的方法是利用集合对值进行排序：

Map<String, Long> map = new HashMap<String, Long>();
// populate with data to sort on Value
// use datastructure designed for sorting

Queue queue = new PriorityQueue( map.size(), new MapComparable() );
queue.addAll( map.entrySet() );

// get a sorted map
LinkedHashMap<String, Long> linkedMap = new LinkedHashMap<String, Long>();

for (Map.Entry<String, Long> entry; (entry = queue.poll())!=null;) {
    linkedMap.put(entry.getKey(), entry.getValue());
}

public static class MapComparable implements Comparator<Map.Entry<String, Long>>{

  public int compare(Entry<String, Long> e1, Entry<String, Long> e2) {
    return e1.getValue().compareTo(e2.getValue());
  }
}

如果您的Map值实现Comparable（例如String），那么这应该会起作用

Map<Object, String> map = new HashMap<Object, String>();
// Populate the Map
List<String> mapValues = new ArrayList<String>(map.values());
Collections.sort(mapValues);

如果映射值本身没有实现Comparable，但您有一个Comparable实例可以对它们进行排序，请将最后一行替换为：

Collections.sort(mapValues, comparable);