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);
    }

我建议使用Arrays.sort，而不是像某些人那样使用Collections.ort。实际上Collections.ort的作用是这样的：

public static <T extends Comparable<? super T>> void sort(List<T> list) {
    Object[] a = list.toArray();
    Arrays.sort(a);
    ListIterator<T> i = list.listIterator();
    for (int j=0; j<a.length; j++) {
        i.next();
        i.set((T)a[j]);
    }
}

它只调用列表上的array，然后使用Arrays.sort。这样，所有映射条目将被复制三次：一次从映射复制到临时列表（无论是LinkedList还是ArrayList），然后复制到临时数组，最后复制到新映射。

我的解决方案省略了这一步，因为它不会创建不必要的LinkedList。以下是代码，通用友好，性能最佳：

public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue(Map<K, V> map) 
{
    @SuppressWarnings("unchecked")
    Map.Entry<K,V>[] array = map.entrySet().toArray(new Map.Entry[map.size()]);

    Arrays.sort(array, new Comparator<Map.Entry<K, V>>() 
    {
        public int compare(Map.Entry<K, V> e1, Map.Entry<K, V> e2) 
        {
            return e1.getValue().compareTo(e2.getValue());
        }
    });

    Map<K, V> result = new LinkedHashMap<K, V>();
    for (Map.Entry<K, V> entry : array)
        result.put(entry.getKey(), entry.getValue());

    return result;
}

Geeks For Geeks对HashMap按值排序

Input : Key = Math, Value = 98
    Key = Data Structure, Value = 85
    Key = Database, Value = 91
    Key = Java, Value = 95
    Key = Operating System, Value = 79
    Key = Networking, Value = 80

Output : Key = Operating System, Value = 79
         Key = Networking, Value = 80
         Key = Data Structure, Value = 85
         Key = Database, Value = 91
         Key = Java, Value = 95
         Key = Math, Value = 98
Solution: The idea is to store the entry set in a list and sort the list on the basis of values. Then fetch values and keys from the list and put them in a new hashmap. Thus, a new hashmap is sorted according to values.
Below is the implementation of the above idea: 




// Java program to sort hashmap by values
import java.util.*;
import java.lang.*;
 
public class GFG {
 
    // function to sort hashmap by values
    public static HashMap<String, Integer> sortByValue(HashMap<String, Integer> hm)
    {
        // Create a list from elements of HashMap
        List<Map.Entry<String, Integer> > list =
               new LinkedList<Map.Entry<String, Integer> >(hm.entrySet());
 
        // Sort the list
        Collections.sort(list, new Comparator<Map.Entry<String, Integer> >() {
            public int compare(Map.Entry<String, Integer> o1,
                               Map.Entry<String, Integer> o2)
            {
                return (o1.getValue()).compareTo(o2.getValue());
            }
        });
         
        // put data from sorted list to hashmap
        HashMap<String, Integer> temp = new LinkedHashMap<String, Integer>();
        for (Map.Entry<String, Integer> aa : list) {
            temp.put(aa.getKey(), aa.getValue());
        }
        return temp;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        HashMap<String, Integer> hm = new HashMap<String, Integer>();
 
        // enter data into hashmap
        hm.put("Math", 98);
        hm.put("Data Structure", 85);
        hm.put("Database", 91);
        hm.put("Java", 95);
        hm.put("Operating System", 79);
        hm.put("Networking", 80);
        Map<String, Integer> hm1 = sortByValue(hm);
 
        // print the sorted hashmap
        for (Map.Entry<String, Integer> en : hm1.entrySet()) {
            System.out.println("Key = " + en.getKey() +
                          ", Value = " + en.getValue());
        }
    }
}
Output
Key = Operating System, Value = 79
Key = Networking, Value = 80
Key = Data Structure, Value = 85
Key = Database, Value = 91
Key = Java, Value = 95
Key = Math, Value = 98

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);

以下是Java8的代码，使用算盘

Map<String, Integer> map = N.asMap("a", 2, "b", 3, "c", 1, "d", 2);
Map<String, Integer> sortedMap = Stream.of(map.entrySet()).sorted(Map.Entry.comparingByValue()).toMap(e -> e.getKey(), e -> e.getValue(),
    LinkedHashMap::new);
N.println(sortedMap);
// output: {c=1, a=2, d=2, b=3}

声明：我是算盘通用的开发者。

因为地图是无序的要对其进行排序，我们可以执行以下操作

Map<String, String> map= new TreeMap<String, String>(unsortMap);

您应该注意，与哈希映射不同，树映射保证其元素将按升序键排序。