面试题：HashMap排序

题目

已知一个HashMap<Integer，User>集合， User有name（String）和age（int）属性。请写一个方法实现对HashMap的排序功能，该方法接收HashMap<Integer，User>为形参，返回类型为HashMap<Integer，User>，要求对HashMap中的User的age正序进行排序。排序时key=value键值对不得拆散。

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我的代码如下

User类

package com.pojo;

public class User {
    private String name;
    private int age;

    public User() {
    }

    public User(String name, int age) {
        this.name = name;
        this.age = age;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    @Override
    public String toString() {
        return "User{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
}

业务代码

package com.collection.list;

import com.pojo.User;

import java.util.*;

public class ListDemo {
    public static void main(String[] args) {
        HashMap<Integer, User> hashMap = new HashMap<Integer, User>(){
            {
                put(1, new User("小明", 12));
                put(2, new User("小张", 16));
                put(3, new User("小赵", 11));
                put(4, new User("小黄", 10));
                put(5, new User("小王", 10));
                put(6, new User("小胡", 11));
                put(7, new User("小梦", 12));
            }
        };

        hashMap = sortMap(hashMap);
        System.out.println(hashMap);
    }

    private static HashMap<Integer, User> sortMap(HashMap<Integer, User> hashMap) {

        TreeSet<Map.Entry<Integer, User>> entries1 = new TreeSet<Map.Entry<Integer, User>>(new Comparator<Map.Entry<Integer, User>>() {
            @Override
            public int compare(Map.Entry<Integer, User> o1, Map.Entry<Integer, User> o2) {
                //一定要注意避免数据丢失，假如写成return o1.getValue().getAge() > o2.getValue().getAge()就会造成数据丢失
                if (o1.getValue().getAge() > o2.getValue().getAge()) {
                    return -1;
                } else {
                    return 1;
                }
            }
        });
        entries1.addAll(hashMap.entrySet());

        LinkedHashMap linkedHashMap = new LinkedHashMap<Integer, User>();
        for (Map.Entry<Integer, User> integerUserEntry : entries1) {
            linkedHashMap.put(integerUserEntry.getKey(), integerUserEntry.getValue());
        }

        return linkedHashMap;
    }
}

--完--