Java - 对嵌套List针对某index进行排序

阳光穿透心脏的1/2处 2023-02-18 10:56 10阅读 0赞

示例：  
\[  
\[3,4,\[name=“list1”,age=1\]\],  
\[2,5,\[name=“list2”,age=2\]\],  
\[1,6\[name=“list3”,age=3\]\]  
\]

我们想针对任意的index进行排序，其中**index=2为Student实例**。  
排序之后输出的结果还是嵌套List,只是输出的顺序发生变化。  
本方法使用`TreeMap`来完成排序操作。

--------------------

### 1、对索引=0，1 进行排序 ###

import java.util.*;
    
    public class Student {
        
        private String name;
        private int age;
    
        public Student() {
        
        }
    
        public Student(String name, int age) {
        
            this.name = name;
            this.age = age;
        }
    
        public static void main(String[] args) {
        
            List<List<Object>> list = new ArrayList<>();
    
    
            Student s1 = new Student("Jerry",1);
            Student s2 = new Student("Tom",2);
            Student s3 = new Student("Jack",3);
    
            List<Object> l1 = new ArrayList<>();
            l1.add(3);
            l1.add(4);
            l1.add(s1);
    
            List<Object> l2 = new ArrayList<>();
            l2.add(2);
            l2.add(5);
            l2.add(s2);
    
            List<Object> l3 = new ArrayList<>();
            l3.add(1);
            l3.add(6);
            l3.add(s3);
    
            list.add(l1);
            list.add(l2);
            list.add(l3);
    
            //调用
            List<List<Object>> lists = Student.OrderByIndex(list, 0,false);
    
            System.out.println(lists.toString());
    
    
        }
    
        public static List<List<Object>> OrderByIndex(List<List<Object>> list,int index,boolean reverse){
        
            List<List<Object>> new_list = new ArrayList<>();
    
            Map<Integer,List<Object>> map = null;
            if (reverse){
        
                map = new TreeMap<>(
                        new Comparator<Integer>() {
        
                            @Override
                            public int compare(Integer o1, Integer o2) {
        
                                //降序排序
                                return o2.compareTo(o1);
                            }
                        }
                );
            }else{
        
                map = new TreeMap<>(
                        new Comparator<Integer>() {
        
                            @Override
                            public int compare(Integer o1, Integer o2) {
        
                                //顺序排序
                                return o1.compareTo(o2);
                            }
                        }
                );
            }
    
            for (int i = 0; i < list.size(); i++) {
        
                Integer temp = (Integer) list.get(i).get(index);
                map.put(temp,list.get(i));
            }
    
            Set<Integer> keySet = map.keySet();
            Iterator<Integer> iterator = keySet.iterator();
    
            while(iterator.hasNext()){
        
                Integer key = iterator.next();
                new_list.add(map.get(key));
            }
            return new_list;
        }
    
        @Override
        public String toString() {
        
            return "age=" + this.age + " and name=" + this.name;
        }
    }

运行结果：

[[1, 6, age=3 and name=Jack], [2, 5, age=2 and name=Tom], [3, 4, age=1 and name=Jerry]]

### 1、对索引=2 进行排序 ###

索引2为Student的实例，需要在方法体中进行改进

import java.util.*;
    
    public class Student {
        
        private String name;
        private int age;
    
        public Student() {
        
        }
    
        public Student(String name, int age) {
        
            this.name = name;
            this.age = age;
        }
    
        public int getAge() {
        
            return age;
        }
    
        public static void main(String[] args) {
        
            List<List<Object>> list = new ArrayList<>();
    
    
            Student s1 = new Student("Jerry",1);
            Student s2 = new Student("Tom",2);
            Student s3 = new Student("Jack",3);
    
            List<Object> l1 = new ArrayList<>();
            l1.add(3);
            l1.add(4);
            l1.add(s1);
    
            List<Object> l2 = new ArrayList<>();
            l2.add(2);
            l2.add(5);
            l2.add(s2);
    
            List<Object> l3 = new ArrayList<>();
            l3.add(1);
            l3.add(6);
            l3.add(s3);
    
            list.add(l1);
            list.add(l2);
            list.add(l3);
    
            //调用
            List<List<Object>> lists = Student.OrderByIndex(list, 2,false);
    
            System.out.println(lists.toString());
    
    
        }
    
        public static List<List<Object>> OrderByIndex(List<List<Object>> list,int index,boolean reverse){
        
            List<List<Object>> new_list = new ArrayList<>();
    
            Map<Integer,List<Object>> map = null;
            if (reverse){
        
                map = new TreeMap<>(
                        new Comparator<Integer>() {
        
                            @Override
                            public int compare(Integer o1, Integer o2) {
        
                                //降序排序
                                return o2.compareTo(o1);
                            }
                        }
                );
            }else{
        
                map = new TreeMap<>(
                        new Comparator<Integer>() {
        
                            @Override
                            public int compare(Integer o1, Integer o2) {
        
                                //顺序排序
                                return o1.compareTo(o2);
                            }
                        }
                );
            }
    
            for (int i = 0; i < list.size(); i++) {
        
            	//获取Student实例
                Student temp = (Student) list.get(i).get(index);
                //key为Student的年龄
                map.put(temp.getAge(),list.get(i));
            }
    
            Set<Integer> keySet = map.keySet();
            Iterator<Integer> iterator = keySet.iterator();
    
            while(iterator.hasNext()){
        
                Integer key = iterator.next();
                new_list.add(map.get(key));
            }
            return new_list;
        }
    
        @Override
        public String toString() {
        
            return "age=" + this.age + " and name=" + this.name;
        }
    }

运行结果：

[[3, 4, age=1 and name=Jerry], [2, 5, age=2 and name=Tom], [1, 6, age=3 and name=Jack]]