使用唯一值合并列表或追加列表中的字段

如何解决使用唯一值合并列表或追加列表中的字段

Label

我必须以最终结果这样的方式合并这两个列表

struct

尝试的解决方案：

• 获取两个列表的 cityName 并且 listA 是否包含 listB 如果为 true 然后在 listB 上运行 for 循环并找到相同的 cityName 并尝试将 userNumber 附加到 listA。

上面的解决方案成功了一半，不确定它的效率如何。请帮忙。

解决方法

您需要使用字段 equals() 和 hashcode() 覆盖 POJO 类中的 country 和 cityName 方法。 那么

1. 将所有元素添加到名为 finalList 的列表中。

2. 迭代 finalList 中的每个元素并将新元素添加到 map 中。

3. 如果在地图中发现重复元素，则将地图中对象的 userNumber 与重复对象的元素合并。并添加回地图。

4. 重复第 2 步和第 3 步，直到迭代完成。

5. 将 map's 键转换为 finalList

示例：

import java.util.*;
import java.util.stream.IntStream;

class MyClass{
    private String country;
    private int[] userNumber;
    private String cityName ;
    MyClass(String country,int[] userNumber,String cityName){
        this.country = country;
        this.userNumber = userNumber;
        this.cityName = cityName;
    }

    public String getCountry() {
        return country;
    }

    public void setCountry(String country) {
        this.country = country;
    }

    public int[] getUserNumber() {
        return userNumber;
    }

    public void setUserNumber(int[] userNumber) {
        this.userNumber = userNumber;
    }

    public String getCityName() {
        return cityName;
    }

    public void setCityName(String cityName) {
        this.cityName = cityName;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        MyClass myClass = (MyClass) o;
        return country.equals(myClass.country) &&
                cityName.equals(myClass.cityName);
    }

    @Override
    public int hashCode() {
        return Objects.hash(country,cityName);
    }
}

public class Test {

    public static void main(String args[]){
        List<MyClass> listA = new ArrayList();
        List<MyClass> finalList =new ArrayList();
                listA.add(new MyClass("BE",new int[]{12345},"Brussels"));
        listA.add(new MyClass("BE","Bruges"));
        listA.add(new MyClass("DE",new int[]{54645},"Frankfurt"));
        List<MyClass> listB = new ArrayList();
        listB.add(new MyClass("BE",new int[]{34356},"Brussels"));
        listB.add(new MyClass("BE",new int[]{64325},"Namur"));
        finalList.addAll(listA);
        finalList.addAll(listB);
        HashMap<MyClass,MyClass> map = new HashMap<>();
        for(MyClass myClass : finalList){
            MyClass objInMap = map.get(myClass);
            if(objInMap!=null){
                int[] userNum = IntStream.concat(Arrays.stream(objInMap.getUserNumber()),Arrays.stream(myClass.getUserNumber()))
                        .distinct()
                        .toArray();
               objInMap.setUserNumber(userNum);
            }else{
                map.put(myClass,myClass);
            }
        }
        finalList = new ArrayList<>(map.keySet());
        System.out.println(finalList);
    }
}

,

    List<Address> first = new ArrayList<>();
    List<Address> second = new ArrayList<>();
    for (Address address : first) {
        int index = second.indexOf(address);
        if(index >= 0){
           first.getUserNumber().addAll(second.get(index).getUserNumber())
        }
    }

不要忘记覆盖equals方法并忽略equals方法中的userNumber字段。

,

    public static <T> T[] concatAll(T[] first,T[]... rest) {
        int totalLength = first.length;
        for (T[] array : rest) {
            totalLength += array.length;
        }
        T[] result = Arrays.copyOf(first,totalLength);
        int offset = first.length;
        for (T[] array : rest) {
            System.arraycopy(array,result,offset,array.length);
            offset += array.length;
        }
        return result;
    }

    public static void main(String[] args) {
        List<MyClass> listA = new ArrayList();
        List<MyClass> finalList = new ArrayList();
        listA.add(new MyClass("BE",new Integer[]{12345},new Integer[]{54645},new Integer[]{34356},new Integer[]{64325},"Namur"));
        finalList.addAll(listA);
        finalList.addAll(listB);
        List<MyClass> merged = new ArrayList<>();
        finalList.parallelStream().collect(Collectors.groupingBy(cl -> (cl.getCountry() + cl.getCityName()),Collectors.toList()))
                .forEach((name,item) -> {
                    item.stream().reduce((a,b) -> new MyClass(a.getCountry(),concatAll(a.getUserNumber(),b.getUserNumber()),a.getCityName())).ifPresent(merged::add);

                });
        System.out.println(merged);
    }
}

class MyClass {
    private String country;
    private Integer[] userNumber;
    private String cityName;

    MyClass(String country,Integer[] userNumber,String cityName) {
        this.country = country;
        this.userNumber = userNumber;
        this.cityName = cityName;
    }

    public String getCountry() {
        return country;
    }

    public void setCountry(String country) {
        this.country = country;
    }

    public Integer[] getUserNumber() {
        return userNumber;
    }

    public void setUserNumber(Integer[] userNumber) {
        this.userNumber = userNumber;
    }

    public String getCityName() {
        return cityName;
    }

    public void setCityName(String cityName) {
        this.cityName = cityName;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        MyClass myClass = (MyClass) o;
        return country.equals(myClass.country) &&
                cityName.equals(myClass.cityName);
    }

    @Override
    public int hashCode() {
        return Objects.hash(country,cityName);
    }
Maybe,It is a great way.