Java提高(5)---map集合排序

时间:2024-07-15 19:07:38

map集合排序

这篇文章讲的不仅仅是map排序,比如把对象按某一属性排序,它都可以解决这些问题。

比如,有N个对象,每个对象有个属性就是成绩,成绩分:优秀,良好,合格。那我们如何按照成绩的好坏进行排序呢,下面请看代码。

1.people对象

package com.test;

/*people对象其实很简单,就提供了三个属性*/

class People {

    private String Name;   //姓名
private String Score; //成绩
private String id; //学号 public String getName() {
return Name;
} public void setName(String name) {
Name = name;
} public String getScore() {
return Score;
} public void setScore(String score) {
Score = score;
} public String getId() {
return id;
} public void setId(String id) {
this.id = id;
} public People(String name, String score, String id) {
super();
Name = name;
Score = score;
this.id = id;
}
}

2.主要方法

package com.test;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;
import java.util.TreeMap; /**
* Map进行多条件排序输出
* 成绩具有优秀,合格,不合格好吃属性。
* 入口Map
* 首先按照优秀,合格,不合格排序
* 然后按照人名的标志Id排序
* 出口Map
*
*
*/
public class MainSort { /**
* 准备参数,创建对象
* @return
*/
private static Map<String, People> getPeopleMap() {
Map<String,People> PeopleMap = new TreeMap<>(); // 创建对象
People b = new People("小明" , "优秀", "b");
People a = new People("小红" , "合格", "a");
People c = new People("丁丁" , "合格", "c");
People d = new People("冬冬" , "良好", "d");
People e = new People("小黄" , "优秀", "e");
People f = new People("小李" , "良好", "f");
People g = new People("小钟" , "优秀", "g"); // 添加乱序key值,把对象放入map集合
PeopleMap.put("xniem", b);
PeopleMap.put("akjd", a);
PeopleMap.put("uioo", c);
PeopleMap.put("qw84", d);
PeopleMap.put("584sdf'", e);
PeopleMap.put("4aisdf", f);
PeopleMap.put("458jsf", g); return PeopleMap;
} /**
* 循环打印Map
*/
private static void show(Map<String, People> PeopleMap) {
// 循环Map 这个打印肯定是无序的,也不是按放入的先后顺序
for (Map.Entry<String, People> PeopleOneMap : PeopleMap.entrySet()) {
People People = PeopleOneMap.getValue();
System.out.println(People.getName() + " " + People.getScore()+ " " + People.getId() );
}
} /*
* 由于List能够直接使用Collections进行排序
* 但是Map不行。
* 这边所做的操作就是先将Map--》List
* 然后对List进行排序
* 然后在讲List--》转换成LinkedHashMap
*
*/
public static Map<String, People> sortMapByValue(Map<String, People> PeopleMap) {
if (PeopleMap == null || PeopleMap.isEmpty()) {
return null;
}
// LinkedHashMap是有序的、或者TreeMap都是有序的(这里只能用LinkedHashMap)
Map<String, People> sortedMap = new LinkedHashMap<String, People>(); /* Set set=PeopleMap.entrySet(); PeopleMap.entrySet()返回的是一个set集合
* 再讲ArrayList(Collection<? extends E> c) 可以放collection,set集合是其子类,map不行哦
* 这步就是把map集合转为ArrayList集合
*/ List<Map.Entry<String, People>> entryList = new ArrayList<Map.Entry<String, People>>(PeopleMap.entrySet()); //这步是关键,进过这步之后,entryList已经是个有序的ArrayList集合了
Collections.sort(entryList, new MapValueComparator());
//通过迭代器取出
Iterator<Map.Entry<String, People>> iter = entryList.iterator();
// Map.Entry<String, People>,就是包装了一个map节点,这个节点封装了key,value值,以及别的值(比如hashmap中哈希码和next指针)
Map.Entry<String, People> tmpEntry = null;
while (iter.hasNext()) {
tmpEntry = iter.next();
sortedMap.put(tmpEntry.getKey(), tmpEntry.getValue());
}
return sortedMap;
} /**
* 主方法
*
*/
public static void main(String[] args) {
// 获取Map
Map<String,People> PeopleMap = getPeopleMap();
// 打印未排序的Map
show(PeopleMap);
System.out.println("-----------before-----------");
// 打印排序完了的Map
show(MainSort.sortMapByValue(PeopleMap));
System.out.println("-----------after------------");
}
}

3.Comparator方法

package com.test;

import java.util.Comparator;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry; class MapValueComparator implements Comparator<Map.Entry<String, People>> { @Override
public int compare(Entry<String, People> o1, Entry<String, People> o2) { // 获取比较的两个对象
People People1 = o1.getValue();
People People2 = o2.getValue(); // 将成绩映射成具有比较关系的字符1、2、3
Map<String,Integer> tasteLev = new HashMap<>();
tasteLev.put("优秀", 1);
tasteLev.put("良好", 2);
tasteLev.put("合格", 3); int cr = 0;
// 判断成绩
int a = tasteLev.get(People2.getScore())-tasteLev.get(People1.getScore());
if (a!=0) {
cr = (a>0) ? -1 : 2;
} else { /*其实上面就可以按成绩优秀,良好,合格排序了,
*在做一步目的,就是在成绩相同的情况下,在按照学号进行排序
*/ // 按照对应的Id排序
a = People2.getId().compareTo(People1.getId());
if (a!=0) {
cr = (a>0)? -2 : 1;
}
}
/* 注意上面对一个返回值对应的就是形成比较层次
* 上层
* --> 2
* --> -1
* 下层
* --> 1
* --> -2
*/
return cr;
}
}

最后我们再来看后台输出

Java提高(5)---map集合排序