学步园

import java.util.Iterator;

import java.util.Set;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) {

        Set set = new TreeSet();
        set.add(new Pair("me", "1000"));

        set.add(new Pair("and", "4000"));
        set.add(new Pair("you", "3000"));

        set.add(new Pair("food", "10000"));
        set.add(new Pair("hungry", "5000"));

        set.add(new Pair("later", "6000"));
        set.add(new Pair("myself", "1000"));

        for (Iterator i = set.iterator(); i.hasNext();)

            System.out.println(i.next());
    }
}

class Pair implements Comparable {
    private final String name;
    private final int number;

    public Pair(String name, int number) {
        this.name = name;
        this.number = number;
    }

    public Pair(String name, String number) throws NumberFormatException {
        this.name = name;
        this.number = Integer.parseInt(number);

    }

    public int compareTo(Object o) {
        if (o instanceof Pair) {
            int cmp = Double.compare(number, ((Pair) o).number);
            if (cmp != 0) {
                return cmp;
            }
            return name.compareTo(((Pair) o).name);
        }

        throw new ClassCastException("Cannot compare Pair with "
                + o.getClass().getName());

    }

    public String toString() {
        return name + ' ' + number;
    }
}

类似的C++代码：

typedef pair<string, int> PAIR;

上面方法的实质意义是：将Map结构中的键值对（Map.Entry）封装成一个自定义的类（结构），或者直接用Map.Entry类。自定义类知道自己应该如何排序，也就是按值排序，具体为自己实现Comparable接口或构造一个Comparator对象，然后不用Map结构而采用有序集合（SortedSet,
TreeSet是SortedSet的一种实现)，这样就实现了Map中sort by value要达到的目的。就是说，不用Map，而是把Map.Entry当作一个对象，这样问题变为实现一个该对象的有序集合或对该对象的集合做排序。既可以用SortedSet，这样插入完成后自然就是有序的了，又或者用一个List或数组，然后再对其做排序（Collections.sort()
or Arrays.sort()）。

Encapsulate the information in its own class. Either implement
Comparable and write rules for the natural ordering or write a
Comparator based on your criteria. Store the information in a sorted
collection, or use the Collections.sort() method.

方法二：

You can also use the following code to sort by value:

public static Map sortByValue(Map map)
{

        List list = new LinkedList(map.entrySet());
        Collections.sort(list, new Comparator() {

            public int compare(Object o1, Object o2) {
                return ((Comparable) ((Map.Entry) (o1)).getValue())
                        .compareTo(((Map.Entry) (o2)).getValue());

            }
        });
        Map result = new LinkedHashMap();

        for (Iterator it = list.iterator(); it.hasNext();) {
            Map.Entry entry = (Map.Entry) it.next();
            result.put(entry.getKey(), entry.getValue());
        }
        return result;
    }

    public static Map sortByValue(Map map, final boolean reverse) {
        List list = new LinkedList(map.entrySet());
        Collections.sort(list, new Comparator() {

            public int compare(Object o1, Object o2) {
                if (reverse) {
                    return -((Comparable) ((Map.Entry) (o1)).getValue())
                            .compareTo(((Map.Entry) (o2)).getValue());
                }
                return ((Comparable) ((Map.Entry) (o1)).getValue())
                        .compareTo(((Map.Entry) (o2)).getValue());
            }
        });

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

        Map map = new HashMap();
        map.put("a", 4);
        map.put("b", 1);
        map.put("c", 3);
        map.put("d", 2);
        Map sorted = sortByValue(map);
        System.out.println(sorted);
// output : {b=1, d=2, c=3, a=4}

或者还可以这样：
Map map = new HashMap();

        map.put("a", 4);
        map.put("b", 1);
        map.put("c", 3);
        map.put("d", 2);

        Set<Map.Entry<String, Integer>> treeSet = new TreeSet<Map.Entry<String, Integer>>(
                new Comparator<Map.Entry<String, Integer>>() {
                    public int compare(Map.Entry<String, Integer> o1,
                            Map.Entry<String, Integer> o2) {
                        Integer d1 = o1.getValue();
                        Integer d2 = o2.getValue();
                        int r = d2.compareTo(d1);

                        if (r != 0)
                            return r;
                        else
                            return o2.getKey().compareTo(o1.getKey());
                    }

                });
        treeSet.addAll(map.entrySet());
        System.out.println(treeSet);
        // output : [a=4, c=3, d=2, b=1]

另外，Groovy 中实现 sort map by value，当然本质是一样的，但却很简洁 : 
用 groovy 中 map 的 sort 方法（需要 groovy 1.6），

def result = map.sort(){ a, b ->  
            b.value.compareTo(a.value)
        }

如：

 ["a":3,"b":1,"c":4,"d":2].sort{ a,b -> a.value - b.value }
 结果为： [b:1, d:2, a:3, c:4]

Python中也类似：

h = {"a":2,"b":1,"c":3}
i = h.items() // i = [('a', 2), ('c', 3), ('b', 1)]
i.sort(lambda (k1,v1),(k2,v2): cmp(v2,v1) ) // i = [('c', 3), ('a', 2), ('b', 1)]