TreeSet按value排序

今天学习到TreeSet，但是它是按照key的值来排序的，那如果我们想要按照value的值排序呢？这个问题我上网看了好久，终于找到一个比较易懂的例子：

例：（统计输入数字的个数）

话不多说，看代码就懂

import java.util.*;

import java.util.Scanner;



public class CountNum {

  public static void main(String[] args){



    HashMap<Integer, Integer> hashMap = new HashMap<Integer,Integer>();

	Scanner input = new Scanner(System.in);

	int num = input.nextInt();

    while (num != 0) 

	{

		if(hashMap.get(num) == null)

		{

		  hashMap.put(num,1);

		}

		else

		{

			int value = hashMap.get(num).intValue();

			value++;

			hashMap.put(num,value);

		}

		num = input.nextInt();

	}



    // Create a tree map from the hash map

    TreeMap<Integer, Integer> treeMap =

      new TreeMap<Integer, Integer>(hashMap);



	Set<Map.Entry<Integer,Integer>> entrySet = treeMap.entrySet();



	ArrayList<Count> list = new ArrayList<Count>();

	for(Map.Entry<Integer,Integer> entry:entrySet){

		Count cc = new Count(entry.getKey(),entry.getValue());

		list.add(cc);

	}

	Collections.sort(list);

/*

	for(int i = 0; i< list.size();i++){

		System.out.println(list.get(i).getKey() + "\t" + list.get(i).getValue());

	}

*/



//找出多个最大值

	int i = 0;

	for(; i< list.size()-1;i++){

		if(list.get(i).getValue() > list.get(i+1).getValue())

		{

			System.out.println(list.get(i).getKey() + "\t" + list.get(i).getValue());

			break;

		}

		else if(list.get(i).getValue() == list.get(i+1).getValue() && i+1 < list.size())

		{

			System.out.println(list.get(i).getKey() + "\t" + list.get(i).getValue());



		}

	}

	if(i == list.size() -1 )

		System.out.println(list.get(i).getKey() + "\t" + list.get(i).getValue());



  }

}



class Count implements Comparable {

  int key;

  int value;



  public int getKey(){

	  return key;

  }

  public int getValue(){

	  return value;

  }

  public Count(int key, int value) {

    this.key = key;

    this.value = value;

  }



  public int compareTo(Object o) {

    return -(value - ((Count)o).value);

  }



  public boolean equals(Object o) {

    return value == ((Count)o).value;

  }

}