Java 排序整合篇

Java 排序整合篇

IILee 24 2022-08-16

前言

Java代码的编写过程中,不可避免会使用到排序功能,本篇博文将常见的数据结构排序罗列出来,供大家参考使用。

正文

List<String>List<Integer>List<Double > 排序

升序:

List<String> stringList = Arrays.asList("d","e","a","b");
Collections.sort(stringList);   //对stringList升序

降序:

List<String> stringList = Arrays.asList("d","e","a","b");
Collections.sort(stringList);     //对stringList降序
Collections.reverse(stringList);  //翻转

Map 根据 key 排序

使用 TreeMap

TreeMap 会根据 keyput 进去的进行升序排序。

升序:

Map<Integer,String> treeMap = new TreeMap<>();
        treeMap.put(1,"111");
        treeMap.put(8,"888");
        treeMap.put(4,"444");
        treeMap.put(2,"222");
System.out.println(treeMap);  //输出{1=111, 2=222, 4=444, 8=888}

降序:

Map<Integer,String> treeMap = new TreeMap<>((o1,o2)->{
            return o2.compareTo(o1);
        });
        treeMap.put(1,"111");
        treeMap.put(8,"888");
        treeMap.put(4,"444");
        treeMap.put(2,"222");
 System.out.println(treeMap);输出{8=888, 4=444, 2=222, 1=111}

注:不推荐用 Double 作为 Mapkey,由于 Map 的底层 hashcode 的实现,会导致取出数据时有时有问题。如果可以确定小数点后的位数,比如小数点后只有两位小数,可以乘以 100 化为 Integer 作为 key,要取出来时除以 10 再取 0

Map 根据 value 排序

value 降序:

Map<String, Double> budget = new HashMap<>();
        budget.put("普通账户", 1000.00);
        budget.put("信用账户", 2000.00);
        budget.put("期权账户",500.00);
        budget.put("理财账户", 4000.00);
//jdk8通过stream实现对Map降序排序
   Map<String, Double> sorted = budget
                .entrySet()
                .stream()
                .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e2, LinkedHashMap::new));
//sorted为{理财账户=4000.0, 信用账户=2000.0, 普通账户=1000.0, 期权账户=500.0}

key 降序并只要前 2 个:

Map<String, Double> budget = new HashMap<>();
        budget.put("普通账户", 1000.00);
        budget.put("信用账户", 2000.00);
        budget.put("期权账户",500.00);
        budget.put("理财账户", 4000.00);

//jdk8通过stream实现对Map降序排序
   Map<String, Double> sorted = budget
                .entrySet()
                .stream()
                .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
                .limit(2)
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e2, LinkedHashMap::new));
//sorted为{理财账户=4000.0, 信用账户=2000.0}

value 升序:

Map<String, Double> budget = new HashMap<>();
        budget.put("普通账户", 1000.00);
        budget.put("信用账户", 2000.00);
        budget.put("期权账户",500.00);
        budget.put("理财账户", 4000.00);

//jdk8通过stream实现对Map升序排序
        Map<String, Double> sorted = budget
                .entrySet()
                .stream()
                .sorted(Map.Entry.comparingByValue(Comparator.naturalOrder()))
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e2, LinkedHashMap::new));
//sorted为{期权账户=500.0, 普通账户=1000.0, 信用账户=2000.0, 理财账户=4000.0}

List<Map> 根据 map 里某一 value 排序

匿名方式:

List<Map<String, Object>> resultlist = new ArrayList<>();
        Map<String, Object> map1 = new HashMap<String, Object>();
        map1.put("id", "1");
        map1.put("name", "张三");
        map1.put("Score", 86.5);
        Map<String, Object> map2 = new HashMap<String, Object>();
        map2.put("id", "2");
        map2.put("name", "李四");
        map2.put("Score", 90.0);
        Map<String, Object> map3 = new HashMap<String, Object>();
        map3.put("id", "3");
        map3.put("name", "王五");
        map3.put("Score", 70.5);

        resultlist.add(map1);
        resultlist.add(map2);
        resultlist.add(map3);

        Collections.sort(resultlist, new Comparator<Map<String, Object>>() {
            @Override
            public int compare(Map<String, Object> o1, Map<String, Object> o2) {
                Double one = (Double) o1.get("Score");
                Double two = (Double) o2.get("Score");
                return one.compareTo(two);  //one.compareTo(two)为升序,two.compareTo(one)为降序
            }
        });

        System.out.println(resultlist);
//输出[{Score=70.5, name=王五, id=3}, {Score=86.5, name=张三, id=1}, {Score=90.0, name=李四, id=2}]

lambda 方式:

List<Map<String, Object>> resultlist = new ArrayList<>();
        Map<String, Object> map1 = new HashMap<String, Object>();
        map1.put("id", "1");
        map1.put("name", "张三");
        map1.put("Score", 86.5);
        Map<String, Object> map2 = new HashMap<String, Object>();
        map2.put("id", "2");
        map2.put("name", "李四");
        map2.put("Score", 90.0);
        Map<String, Object> map3 = new HashMap<String, Object>();
        map3.put("id", "3");
        map3.put("name", "王五");
        map3.put("Score", 70.5);

        resultlist.add(map1);
        resultlist.add(map2);
        resultlist.add(map3);
        Collections.sort(resultlist, (o1, o2) -> {
            Double one =  Double.valueOf(o1.get("Score").toString()) ;
            Double two =  Double.valueOf(o2.get("Score").toString()) ;
            return one.compareTo(two);  //one.compareTo(two)为升序,two.compareTo(one)为降序
        });
        System.out.println(resultlist);
//输出[{Score=70.5, name=王五, id=3}, {Score=86.5, name=张三, id=1}, {Score=90.0, name=李四, id=2}]

List<Object> 根据 Object 的某一属性对 List 进行排序

Student 类:

public class Student {
    private String name;
    private int score;

    public String getName() {
        return name;
    }

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

    public int getScore() {
        return score;
    }

    public void setScore(int score) {
        this.score = score;
    }
}

List<Student> 排序:

List<Student> list = new ArrayList<>();
        Student student1=new Student();
        Student student2=new Student();
        Student student3=new Student();
        student1.setName("张三");
        student1.setScore(92);
        student2.setName("李四");
        student2.setScore(86);
        student3.setName("王五");
        student3.setScore(100);
        list.add(student1);
        list.add(student2);
        list.add(student3);

        System.out.println("排序前:");
        for(int i = 0;i < list.size();i++){
            System.out.println(list.get(i).getName()+":"+list.get(i).getScore());
        }
        System.out.println("--------");
        System.out.println("排序后:");
        Collections.sort(list, new Comparator<Student>() {
            @Override
            public int compare(Student o1, Student o2) {
                int i = o1.getScore() - o2.getScore();
                return i;
            }
        });
        for(int i = 0;i < list.size();i++){
            System.out.println(list.get(i).getName()+":"+list.get(i).getScore());
        }

结果:

排序前:
张三:92
李四:86
王五:100
--------
排序后:
李四:86
张三:92
王五:100

复杂 List<Map<String, List<Map<String, Object>>>> 排序实例

要求:不仅内部要根据 Score 排序,整个 List 也要根据 Score 排序

List<Map<String, List<Map<String, Object>>>> tagsList = new ArrayList<>();

        List<Map<String, Object>> resultlist1 = new ArrayList<>();
        Map<String, List<Map<String, Object>>> listMap1 = new HashMap<>();
        Map<String, Object> map1 = new HashMap<String, Object>();
        map1.put("id", "1");
        map1.put("name", "张一");
        map1.put("Score", 7L);
        Map<String, Object> map2 = new HashMap<String, Object>();
        map2.put("id", "2");
        map2.put("name", "李一");
        map2.put("Score", 100L);
        Map<String, Object> map3 = new HashMap<String, Object>();
        map3.put("id", "3");
        map3.put("name", "王一");
        map3.put("Score", 86L);
        Map<String, Object> map4 = new HashMap<String, Object>();
        map4.put("id", "4");
        map4.put("name", "赵一");
        map4.put("Score", 0L);
        resultlist1.add(map1);
        resultlist1.add(map2);
        resultlist1.add(map3);
        resultlist1.add(map4);
        listMap1.put("1", resultlist1);
        tagsList.add(listMap1);


        List<Map<String, Object>> resultlist2 = new ArrayList<>();
        Map<String, List<Map<String, Object>>> listMap2 = new HashMap<>();
        Map<String, Object> map2_1 = new HashMap<String, Object>();
        map2_1.put("id", "1");
        map2_1.put("name", "张二");
        map2_1.put("Score", 709L);
        Map<String, Object> map2_2 = new HashMap<String, Object>();
        map2_2.put("id", "2");
        map2_2.put("name", "李二");
        map2_2.put("Score", 0L);
        Map<String, Object> map2_3 = new HashMap<String, Object>();
        map2_3.put("id", "3");
        map2_3.put("name", "王二");
        map2_3.put("Score", 456L);
        Map<String, Object> map3_4 = new HashMap<String, Object>();
        map3_4.put("id", "4");
        map3_4.put("name", "赵二");
        map3_4.put("Score", 0L);
        resultlist2.add(map2_1);
        resultlist2.add(map2_1);
        resultlist2.add(map2_3);
        resultlist2.add(map3_4);
        listMap2.put("2", resultlist2);
        tagsList.add(listMap2);

        List<Map<String, Object>> resultlist3 = new ArrayList<>();
        Map<String, List<Map<String, Object>>> listMap3 = new HashMap<>();
        Map<String, Object> map3_1 = new HashMap<String, Object>();
        map3_1.put("id", "1");
        map3_1.put("name", "张三");
        map3_1.put("Score", 9L);
        Map<String, Object> map3_2 = new HashMap<String, Object>();
        map3_2.put("id", "2");
        map3_2.put("name", "李三");
        map3_2.put("Score", 0L);
        Map<String, Object> map3_3 = new HashMap<String, Object>();
        map3_3.put("id", "3");
        map3_3.put("name", "王三");
        map3_3.put("Score", 86L);
        Map<String, Object> map2_4 = new HashMap<String, Object>();
        map2_4.put("id", "4");
        map2_4.put("name", "赵三");
        map2_4.put("Score", 920L);
        resultlist3.add(map3_1);
        resultlist3.add(map3_2);
        resultlist3.add(map3_3);
        resultlist3.add(map2_4);
        listMap3.put("3", resultlist3);
        tagsList.add(listMap3);

        System.out.println("未排序时的tagsList:" + tagsList);

        //先对每一个里的Score降序排序,排序后tagsList里的每一个元素中的List<Map<String, Object>>都是有序的
        for (Map<String, List<Map<String, Object>>> ins : tagsList) {
            for (List<Map<String, Object>> v : ins.values()) {
                Collections.sort(v, (o1, o2) -> {
                    Long one = Long.valueOf(o1.get("Score").toString());
                    Long two = Long.valueOf(o2.get("Score").toString());
                    return two.compareTo(one);
                });
            }
        }
        System.out.println("对每一个Score内部降序排序后的tagsList:" + tagsList);

        Collections.sort(tagsList, (o1, o2) -> {
            Long one = 0L;
            Long two = 0L;
            for (List<Map<String, Object>> v : o1.values()) {
                one = Long.valueOf(v.get(0).get("Score").toString());
                break;
            }
            for (List<Map<String, Object>> v : o2.values()) {
                two = Long.valueOf(v.get(0).get("Score").toString());
                break;
            }
            return two.compareTo(one);
        });
        System.out.println("根据最内层的最大值排序后的tagsList:" + tagsList);

输出结果:

未排序时的tagsList:
[{1=[{Score=7, name=张一, id=1}, {Score=100, name=李一, id=2}, {Score=86, name=王一, id=3}, {Score=0, name=赵一, id=4}]}, 
{2=[{Score=709, name=张二, id=1}, {Score=709, name=张二, id=1}, {Score=456, name=王二, id=3}, {Score=0, name=赵二, id=4}]}, 
{3=[{Score=9, name=张三, id=1}, {Score=0, name=李三, id=2}, {Score=86, name=王三, id=3}, {Score=920, name=赵三, id=4}]}]
    
对每一个Score内部降序排序后的tagsList:
[{1=[{Score=100, name=李一, id=2}, {Score=86, name=王一, id=3}, {Score=7, name=张一, id=1}, {Score=0, name=赵一, id=4}]}, 
{2=[{Score=709, name=张二, id=1}, {Score=709, name=张二, id=1}, {Score=456, name=王二, id=3}, {Score=0, name=赵二, id=4}]}, 
{3=[{Score=920, name=赵三, id=4}, {Score=86, name=王三, id=3}, {Score=9, name=张三, id=1}, {Score=0, name=李三, id=2}]}]
    
根据最内层的最大值排序后的tagsList:
[{3=[{Score=920, name=赵三, id=4}, {Score=86, name=王三, id=3}, {Score=9, name=张三, id=1}, {Score=0, name=李三, id=2}]}, 
{2=[{Score=709, name=张二, id=1}, {Score=709, name=张二, id=1}, {Score=456, name=王二, id=3}, {Score=0, name=赵二, id=4}]}, 
{1=[{Score=100, name=李一, id=2}, {Score=86, name=王一, id=3}, {Score=7, name=张一, id=1}, {Score=0, name=赵一, id=4}]}]

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