Question

我有两个列表List<AreaOfInterest>和List<AreaOfInterestBean>。相应的类是

@Entity
@Data
@Table( name = "interest_area" )
public class AreaOfInterest {

    @Id
    @GeneratedValue( strategy = GenerationType.AUTO )
    @Column( name = "ia_id" )
    private Long id;

    @Column( name = "ia_value" )
    private String value;

    @Column( name = "ia_is_active" )
    private Boolean isActive;

    @Column( name = "ia_created_at" )
    private Calendar createdAt;

    @Column( name = "ia_is_approved" )
    private Boolean isApprovedByAdmin;

    @Column( name = "ia_approved_at" )
    private Calendar approvedAt;

}

和

@Data
public class AreaOfInterestBean {

    private Long id;

    private String value;

    private boolean isSelected;
}

我想检查List<AreaOfInterest>中是否存在List<AreaOfInterestBean>的对象，如果存在，则需要更新List<AreaOfInterestBean>的相应对象。现在，我觉得我必须遍历两个列表才能解决此问题。有什么有效的方法吗？

Answer 1

说实话，我不确定这个过程的复杂性（我明天会尝试解决）：

Set<String> left = nonBeans.stream()
            .map(AreaOfInterest::getValue)
            .collect(Collectors.toSet());

beans.stream()
      .filter(x -> left.contains(x.getValue()))
      .forEach(x -> x.setSelected(true));

在这种情况下，复杂度将为O(n)。

Answer 2

假设：

List<AreaOfInterest> list = /* ... */
List<AreaOfInterestBean> listOfBeans = /* ... */

只需两个简单的迭代即可轻松实现：

for (int i=0; i<listOfBeans.size(); i++) {                // Iterate beans
    AreaOfInterestBean aoiBean = listOfBeans.get(i);      // Get each bean
    for (int j=0; j<list.size(); j++) {                   // Iterate entity
        AreaOfInterest aoi = list.get(j);                 // Get each entity
        if (aoiBean.getId() == aoi.getId()) {             // If they have the same ID
            aoiBean.setValue(aoi.getValue());             // Update the bean value
            break;                                        // Go to the next bean
        }
    }
}

示例：具有以下简化数据：

AreaOfInterest和id和value：1, "Car"，2, "Bike"，3, "Ship"
AreaOfInterestBeans和id和value：5, "Limo"，2, "Scooter"，7, "Yacht"

结果AreaOfInterestBeans将具有ID 2的替换值：Scooter-> Bike。

AreaOfInterestBeans和id和value：5, "Limo"，2, "Bike"，7, "Yacht"

另一种Stream方法：

listOfBeans = listOfBeans.stream()
                  .peek(aoiBean -> list.stream()
                                       .filter(aoi -> aoi.getId() == aoiBean.getId())
                                       .findFirst()
                                       .ifPresent(i -> aoiBean.setValue(i.getValue())))
                  .collect(Collectors.toList());

查找列表的对象是否存在于另一个列表中

2 个答案: