Question

1医院将有多名医生。我需要知道java collection类型（ArrayList,HashMap等），我可以将hospital ID和Doctor对象存储在合适的java collection类型中{{1}等等）。

要求是我应该能够将ArrayList,HashMap存储为HospitalID，将key对象存储为Doctor。

此外，我应该能够为各种Doctor对象提供相同的密钥（因为可能有许多医生为该医院工作）。那么我可以用于这种情况的value集合类型（java等）是什么？

注意：我无法使用ArrayList,HashMap - 因为它需要唯一的ID。

稍后，我应该能够过滤掉所有为特定医院工作的医生（通过搜索它的ID），并显示其记录

Answer 1

如果您将自己限制在标准集合类型之外，则需要使用Map<HospitalId, Set<Doctor>>。如果您可以使用第三方库，那么您要寻找的是“多图”。

不同实现类（HashSet与TreeSet之间的选择等）取决于您打算使用数据结构的方式。

Answer 2

您可以拥有ArrayList个Doctor个对象，然后创建一个HashMap来存储HospitalID作为关键字，ArrayList个医生作为值：

ArrayList<Doctor> a = new ArrayList<Doctor>();
a.add(new Doctor());
// put all the doctors

HashMap<Integer,ArrayList<Doctor>> hMap = new HashMap<Integer,ArrayList<Doctor>>();
Integer hospitalId = new Intger(1);

hMap.put(hospitalId,a);

更新：

添加新医生：

//Take the existing list from the map using hospitalId
ArrayList<Doctor> existingList = hMap.get(hospitalId);

Doctor d = new Doctor();
// add new doctor to existingList
existingList.add(d);

//put the new list again in the map

hMap.put(hospitalId,existingList);

Answer 3

许多医生可以与1个医院id相关。因此，有一对多的映射。我认为你应该使用

地图（医院数据集，Dcotors数组列表）

其中set是hospital_id的集合，并且它是唯一的，Arraylist是医生的集合。

因此，1 hospital_id可以包含医生名单。

Answer 4

您需要查看区别b / w列表和地图，并根据您的要求选择。

<强>列表：

有序集合（也称为序列）。该接口的用户可以精确控制列表中每个元素的插入位置。用户可以通过整数索引（列表中的位置）访问元素，并搜索列表中的元素。

<强>地图

将键映射到值的对象。地图不能包含重复的键;每个键最多可以映射一个值。

如果您的ID也与其他医生相同，那么您可以将ID与Doctor结合使用，并将其作为单个对象保存在ArrayList中。

Answer 5

创建一个名为 Hospital.java

的类

package com.rais.hospital;

/**
 * @author Rais.Alam
 * @project MyFirstProject
 * @date Dec 24, 2012
 */


public class Hospital
{
private Integer hospitalId;
private String hospitalName;
private String hospitalAddress;
private Long contatNumber;
/**
 * @param hospitalId
 * @param hospitalName
 * @param hospitalAddress
 * @param contatNumber
 */



public Hospital(Integer hospitalId, String hospitalName, String hospitalAddress, Long contatNumber)
{
    super();
    this.hospitalId = hospitalId;
    this.hospitalName = hospitalName;
    this.hospitalAddress = hospitalAddress;
    this.contatNumber = contatNumber;
}
/**
 * @param hospitalId
 * @param contatNumber
 */
public Hospital(Integer hospitalId)
{
    super();
    this.hospitalId = hospitalId;

}
/**
 * @return the hospitalId
 */
public Integer getHospitalId()
{
    return hospitalId;
}
/**
 * @param hospitalId the hospitalId to set
 */
public void setHospitalId(Integer hospitalId)
{
    this.hospitalId = hospitalId;
}
/**
 * @return the hospitalName
 */
public String getHospitalName()
{
    return hospitalName;
}
/**
 * @param hospitalName the hospitalName to set
 */
public void setHospitalName(String hospitalName)
{
    this.hospitalName = hospitalName;
}
/**
 * @return the hospitalAddress
 */
public String getHospitalAddress()
{
    return hospitalAddress;
}
/**
 * @param hospitalAddress the hospitalAddress to set
 */
public void setHospitalAddress(String hospitalAddress)
{
    this.hospitalAddress = hospitalAddress;
}
/**
 * @return the contatNumber
 */
public Long getContatNumber()
{
    return contatNumber;
}
/**
 * @param contatNumber the contatNumber to set
 */
public void setContatNumber(Long contatNumber)
{
    this.contatNumber = contatNumber;
}
/* (non-Javadoc)
 * @see java.lang.Object#hashCode()
 */
@Override
public int hashCode()
{
    final int prime = 31;
    int result = 1;
    result = prime * result + ((hospitalId == null) ? 0 : hospitalId.hashCode());

    return result;
}
/* (non-Javadoc)
 * @see java.lang.Object#equals(java.lang.Object)
 */
@Override
public boolean equals(Object obj)
{
    if (this == obj)
        return true;
    if (obj == null)
        return false;
    if (getClass() != obj.getClass())
        return false;
    Hospital other = (Hospital) obj;

    if (!hospitalId.equals(other.hospitalId))
        return false;

    return true;


}




}

创建一个类 Doctor.java

package com.rais.hospital;

/**
 * @author Rais.Alam
 * @project MyFirstProject
 * @date Dec 24, 2012
 */
public class Doctor
{
private Integer id;
private String name;
private String address;
private String department;



/**
 * @param id
 * @param name
 * @param address
 * @param department
 */
public Doctor(Integer id, String name, String address, String department)
{
    super();
    this.id = id;
    this.name = name;
    this.address = address;
    this.department = department;
}
/**
 * @return the id
 */
public Integer getId()
{
    return id;
}
/**
 * @param id the id to set
 */
public void setId(Integer id)
{
    this.id = id;
}
/**
 * @return the name
 */
public String getName()
{
    return name;
}
/**
 * @param name the name to set
 */
public void setName(String name)
{
    this.name = name;
}
/**
 * @return the address
 */
public String getAddress()
{
    return address;
}
/**
 * @param address the address to set
 */
public void setAddress(String address)
{
    this.address = address;
}
/**
 * @return the department
 */
public String getDepartment()
{
    return department;
}
/**
 * @param department the department to set
 */
public void setDepartment(String department)
{
    this.department = department;
}
/* (non-Javadoc)
 * @see java.lang.Object#toString()
 */
@Override
public String toString()
{
    return "Doctor [id=" + id + ", name=" + name + ", address=" + address + ", department=" + department + "]";
}





}

现在在下面描述的客户的帮助下访问所有医生名单 Client.java

package com.rais.hospital;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
/**
 * @author Rais.Alam
 * @project MyFirstProject
 * @date Dec 24, 2012
 */
public class Client
{
private static Map<Hospital, List<Doctor>> repo = new HashMap<Hospital, List<Doctor>>();

/**
 * @param args
 */
public static void main(String[] args)
{

    // Displaying records for Hospital for HospitalA of Boston

    createRepository();

    List<Doctor> lst1 = getDoctorsList(new Hospital(101));

    for (Doctor doctor : lst1)
    {
        System.out.println(doctor);
    }

    System.out.println("==================================");

    // Displaying records for Hospital for HospitalB of Atlanta
    List<Doctor> lst2 = getDoctorsList(new Hospital(201));

    for (Doctor doctor : lst2)
    {
        System.out.println(doctor);
    }

}

public static List<Doctor> getDoctorsList(Hospital hospital)
{

    return repo.get(hospital);

}

public static void createRepository()
{
    Hospital hospital1 = new Hospital(101, "HospitalA", "Street no, 101, Boston", 123456789l);
    Hospital hospital2 = new Hospital(201, "HospitalB", "Street no, 102, Atlanta", 987654321l);

    List<Doctor> list1 = new ArrayList<Doctor>();
    List<Doctor> list2 = new ArrayList<Doctor>();

    list1.add(new Doctor(1011, "Doctor-P", "Boston", "ENT"));
    list1.add(new Doctor(1012, "Doctor-Q", "Boston", "ENT"));
    list1.add(new Doctor(1013, "Doctor-R", "Boston", "ENT"));
    list1.add(new Doctor(1014, "Doctor-S", "Boston", "ENT"));

    list2.add(new Doctor(2011, "Doctor-A", "Atlanta", "Therapist"));
    list2.add(new Doctor(2012, "Doctor-B", "Atlanta", "Therapist"));
    list2.add(new Doctor(2013, "Doctor-C", "Atlanta", "Therapist"));
    list2.add(new Doctor(2014, "Doctor-D", "Atlanta", "Therapist"));

    repo.put(hospital1, list1);
    repo.put(hospital2, list2);

    }

}

您将看到如下所述的输出

Doctor [id=1011, name=Doctor-P, address=Boston, department=ENT]
Doctor [id=1012, name=Doctor-Q, address=Boston, department=ENT]
Doctor [id=1013, name=Doctor-R, address=Boston, department=ENT]
Doctor [id=1014, name=Doctor-S, address=Boston, department=ENT]
==================================
Doctor [id=2011, name=Doctor-A, address=Atlanta, department=Therapist]
Doctor [id=2012, name=Doctor-B, address=Atlanta, department=Therapist]
Doctor [id=2013, name=Doctor-C, address=Atlanta, department=Therapist]
Doctor [id=2014, name=Doctor-D, address=Atlanta, department=Therapist]

Answer 6

听起来你想要Multimap。默认情况下，Collections库没有其中一个，但您可以轻松地从Map和List构建一个：

Map<Hospital, List<Doctor>> = new HashMap<Hospital, LinkedList<Doctor>>();

Answer 7

最好使用**地图＆gt; ** **地图＆gt; **。列出所有在ArrayList中设置的医生。订单权利没有任何优势。 Go with Set它将提供比Map＆gt;中的ArrayList更好的性能。

我应该使用什么java集合类型 - 初学者

7 个答案: