我们只需像这样对地图进行排序

            Map<String, String> unsortedMap = new HashMap<String, String>();

    unsortedMap.put("E", "E Val");
    unsortedMap.put("F", "F Val");
    unsortedMap.put("H", "H Val");
    unsortedMap.put("B", "B Val");
    unsortedMap.put("C", "C Val");
    unsortedMap.put("A", "A Val");
    unsortedMap.put("G", "G Val");
    unsortedMap.put("D", "D Val");

    Map<String, String> sortedMap = new TreeMap<String, String>(unsortedMap);

    System.out.println("\nAfter sorting..");
    for (Map.Entry <String, String> mapEntry : sortedMap.entrySet()) {
        System.out.println(mapEntry.getKey() + " \t" + mapEntry.getValue());

您可以尝试Guava的多功能地图：

TreeMap<Integer, Collection<String>> sortedMap = new TreeMap<>(
        Multimaps.invertFrom(Multimaps.forMap(originalMap), 
        ArrayListMultimap.<Integer, String>create()).asMap());

因此，您将获得从原始值到对应于它们的键集合的映射。即使同一值有多个键，也可以使用此方法。

这还有一个额外的好处，即可以使用Java8进行升序或降序排序

import static java.util.Comparator.comparingInt;
import static java.util.stream.Collectors.toMap;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.stream.Collectors;
import java.util.stream.Stream;

class Utils {
    public static Map<String, Integer> sortMapBasedOnValues(Map<String, Integer> map, boolean descending) {
        int multiplyBy = (descending) ? -1: 1;
        Map<String, Integer> sorted =  map.entrySet().stream()
                .sorted(comparingInt(e -> multiplyBy * e.getValue() ))
                .collect(toMap(
                        Map.Entry::getKey, 
                        Map.Entry::getValue,
                        (a, b) -> { throw new AssertionError();},
                        LinkedHashMap::new
                    ));
        return sorted;
    }
}

从…起http://www.programmersheaven.com/download/49349/download.aspx

private static <K, V> Map<K, V> sortByValue(Map<K, V> map) {
    List<Entry<K, V>> list = new LinkedList<>(map.entrySet());
    Collections.sort(list, new Comparator<Object>() {
        @SuppressWarnings("unchecked")
        public int compare(Object o1, Object o2) {
            return ((Comparable<V>) ((Map.Entry<K, V>) (o1)).getValue()).compareTo(((Map.Entry<K, V>) (o2)).getValue());
        }
    });

    Map<K, V> result = new LinkedHashMap<>();
    for (Iterator<Entry<K, V>> it = list.iterator(); it.hasNext();) {
        Map.Entry<K, V> entry = (Map.Entry<K, V>) it.next();
        result.put(entry.getKey(), entry.getValue());
    }

    return result;
}

如果没有大于地图大小的值，可以使用数组，这应该是最快的方法：

public List<String> getList(Map<String, Integer> myMap) {
    String[] copyArray = new String[myMap.size()];
    for (Entry<String, Integer> entry : myMap.entrySet()) {
        copyArray[entry.getValue()] = entry.getKey();
    }
    return Arrays.asList(copyArray);
}

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