private static List<AddressList> list = new ArrayList<>();怎么改成数组
时间: 2024-01-01 15:10:48 浏览: 60
可以将它改为以下形式:
```
private static AddressList[] list = new AddressList[size];
```
其中,`size`为数组的大小。你需要根据你的实际需求来确定数组的大小。
相关问题
java两个list根据对象的id进行合并成一个对象
在Java中,如果有两个List集合,并且每个集合中的对象都有一个唯一的id属性,我们可以通过这个id属性将两个List合并成一个对象。以下是一个示例代码,展示了如何实现这一操作:
假设我们有两个类:Person和Address,并且我们有两个List集合,一个包含Person对象,另一个包含Address对象。我们希望根据id将这两个List合并成一个包含PersonWithAddress对象的List。
```java
import java.util.*;
import java.util.stream.Collectors;
class Person {
private Integer id;
private String name;
// 构造方法
public Person(Integer id, String name) {
this.id = id;
this.name = name;
}
// Getter方法
public Integer getId() {
return id;
}
public String getName() {
return name;
}
}
class Address {
private Integer id;
private String address;
// 构造方法
public Address(Integer id, String address) {
this.id = id;
this.address = address;
}
// Getter方法
public Integer getId() {
return id;
}
public String getAddress() {
return address;
}
}
class PersonWithAddress {
private Integer id;
private String name;
private String address;
// 构造方法
public PersonWithAddress(Integer id, String name, String address) {
this.id = id;
this.name = name;
this.address = address;
}
// Getter方法
public Integer getId() {
return id;
}
public String getName() {
return name;
}
public String getAddress() {
return address;
}
@Override
public String toString() {
return "PersonWithAddress{id=" + id + ", name='" + name + '\'' + ", address='" + address + '\'' + '}';
}
}
public class MergeListsExample {
public static void main(String[] args) {
List<Person> personList = new ArrayList<>();
personList.add(new Person(1, "Alice"));
personList.add(new Person(2, "Bob"));
personList.add(new Person(3, "Charlie"));
List<Address> addressList = new ArrayList<>();
addressList.add(new Address(1, "New York"));
addressList.add(new Address(2, "Los Angeles"));
addressList.add(new Address(3, "Chicago"));
Map<Integer, Person> personMap = personList.stream().collect(Collectors.toMap(Person::getId, person -> person));
Map<Integer, Address> addressMap = addressList.stream().collect(Collectors.toMap(Address::getId, address -> address));
List<PersonWithAddress> personWithAddressList = personList.stream()
.filter(person -> addressMap.containsKey(person.getId()))
.map(person -> new PersonWithAddress(person.getId(), person.getName(), addressMap.get(person.getId()).getAddress()))
.collect(Collectors.toList());
personWithAddressList.forEach(System.out::println);
}
}
```
在这个示例中,我们首先创建了Person和Address类的对象,并将它们添加到各自的List中。然后,我们使用Stream API将这两个List转换为Map,以便于根据id进行查找。最后,我们遍历personList,查找对应的Address对象,并创建一个新的PersonWithAddress对象,将其添加到最终的List中。
将A,B两个自定义对象类型的List合并为一个Map,其中集合A作为key,集合B作为value,提供java代码
下面是一个简单的Java代码示例,将两个自定义对象类型的List合并为一个Map,其中集合A作为key,集合B作为value:
```java
import java.util.*;
public class MergeListsToMap {
public static void main(String[] args) {
// Create two custom object type lists
List<Person> personList = new ArrayList<>();
personList.add(new Person("Alice", 25));
personList.add(new Person("Bob", 30));
List<Address> addressList = new ArrayList<>();
addressList.add(new Address("123 Main St", "Anytown", "USA"));
addressList.add(new Address("456 Elm St", "Othertown", "USA"));
// Merge the lists into a map
Map<List<Person>, List<Address>> map = new HashMap<>();
map.put(personList, addressList);
// Print out the map
System.out.println(map);
}
}
class Person {
private String name;
private int age;
public Person(String name, int age) {
this.name = name;
this.age = age;
}
public String getName() {
return name;
}
public int getAge() {
return age;
}
}
class Address {
private String street;
private String city;
private String country;
public Address(String street, String city, String country) {
this.street = street;
this.city = city;
this.country = country;
}
public String getStreet() {
return street;
}
public String getCity() {
return city;
}
public String getCountry() {
return country;
}
}
```
在这个示例中,我们创建了两个自定义对象类型的List,一个是Person对象的列表,另一个是Address对象的列表。我们使用HashMap将这两个列表合并为一个Map,其中Person列表作为键,Address列表作为值。最后,我们打印出这个Map来验证它是否正确合并了两个列表。
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