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);
}

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。

迟到。

随着Java-8的出现，我们可以以非常简单/简洁的方式使用流进行数据操作。您可以使用流按值对映射条目进行排序，并创建一个LinkedHashMap，以保留插入顺序迭代。

Eg:

LinkedHashMap sortedByValueMap = map.entrySet().stream()
                .sorted(comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey))     //first sorting by Value, then sorting by Key(entries with same value)
                .collect(LinkedHashMap::new,(map,entry) -> map.put(entry.getKey(),entry.getValue()),LinkedHashMap::putAll);

对于反向排序，请替换：

comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey)

with

comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey).reversed()

HashMap＜String，Long＞的最简单的暴力排序HashMap方法：您可以复制粘贴它，然后这样使用：

public class Test  {
    public static void main(String[] args)  {
        HashMap<String, Long> hashMap = new HashMap<>();
        hashMap.put("Cat", (long) 4);
        hashMap.put("Human", (long) 2);
        hashMap.put("Dog", (long) 4);
        hashMap.put("Fish", (long) 0);
        hashMap.put("Tree", (long) 1);
        hashMap.put("Three-legged-human", (long) 3);
        hashMap.put("Monkey", (long) 2);

        System.out.println(hashMap);  //{Human=2, Cat=4, Three-legged-human=3, Monkey=2, Fish=0, Tree=1, Dog=4}
        System.out.println(sortHashMap(hashMap));  //{Cat=4, Dog=4, Three-legged-human=3, Human=2, Monkey=2, Tree=1, Fish=0}
    }

    public LinkedHashMap<String, Long> sortHashMap(HashMap<String, Long> unsortedMap)  {
        LinkedHashMap<String, Long> result = new LinkedHashMap<>();

        //add String keys to an array: the array would get sorted, based on those keys' values
        ArrayList<String> sortedKeys = new ArrayList<>();
        for (String key: unsortedMap.keySet())  {
            sortedKeys.add(key);
        }

        //sort the ArrayList<String> of keys    
        for (int i=0; i<unsortedMap.size(); i++)  {
            for (int j=1; j<sortedKeys.size(); j++)  {
                if (unsortedMap.get(sortedKeys.get(j)) > unsortedMap.get(sortedKeys.get(j-1))) {
                    String temp = sortedKeys.get(j);
                    sortedKeys.set(j, sortedKeys.get(j-1));
                    sortedKeys.set(j-1, temp);
                }
            }
        }

        // construct the result Map
        for (String key: sortedKeys)  {
            result.put(key, unsortedMap.get(key));
        }

        return result;
    }
}

主要问题。如果您使用第一个答案（Google将您带到这里），请更改比较器以添加等号子句，否则无法按键从sorted_map中获取值：

public int compare(String a, String b) {
        if (base.get(a) > base.get(b)) {
            return 1;
        } else if (base.get(a) < base.get(b)){
            return -1;
        } 

        return 0;
        // returning 0 would merge keys
    }