创建自定义比较器，并在创建新的TreeMap对象时使用它。

class MyComparator implements Comparator<Object> {

    Map<String, Integer> map;

    public MyComparator(Map<String, Integer> map) {
        this.map = map;
    }

    public int compare(Object o1, Object o2) {

        if (map.get(o2) == map.get(o1))
            return 1;
        else
            return ((Integer) map.get(o2)).compareTo((Integer)     
                                                            map.get(o1));

    }
}

在主函数中使用以下代码

    Map<String, Integer> lMap = new HashMap<String, Integer>();
    lMap.put("A", 35);
    lMap.put("B", 75);
    lMap.put("C", 50);
    lMap.put("D", 50);

    MyComparator comparator = new MyComparator(lMap);

    Map<String, Integer> newMap = new TreeMap<String, Integer>(comparator);
    newMap.putAll(lMap);
    System.out.println(newMap);

输出：

{B=75, D=50, C=50, A=35}

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

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;
    }
}

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;
}

要使用Java 8中的新功能实现这一点，请执行以下操作：

import static java.util.Map.Entry.comparingByValue;
import static java.util.stream.Collectors.toList;

<K, V> List<Entry<K, V>> sort(Map<K, V> map, Comparator<? super V> comparator) {
    return map.entrySet().stream().sorted(comparingByValue(comparator)).collect(toList());
}

条目使用给定的比较器按其值排序。或者，如果您的值可以相互比较，则不需要显式比较器：

<K, V extends Comparable<? super V>> List<Entry<K, V>> sort(Map<K, V> map) {
    return map.entrySet().stream().sorted(comparingByValue()).collect(toList());
}

返回的列表是调用此方法时给定映射的快照，因此两者都不会反映对另一个的后续更改。对于地图的实时可迭代视图：

<K, V extends Comparable<? super V>> Iterable<Entry<K, V>> sort(Map<K, V> map) {
    return () -> map.entrySet().stream().sorted(comparingByValue()).iterator();
}

返回的可迭代对象在每次迭代时都会创建给定映射的新快照，因此除非并发修改，否则它将始终反映映射的当前状态。

根据上下文，使用java.util.LinkedHashMap<T>来记住项目在映射中的放置顺序。否则，如果您需要根据值的自然排序对值进行排序，我建议您维护一个单独的List，该List可以通过Collections.sort（）进行排序。