我有两个课程,如下所示。我需要使用这两个类来提取一些东西。
public final class ProcessMetadata {
private final String clientId;
private final String deviceId;
// .. lot of other fields here
// getters here
}
public final class ProcMetadata {
private final String deviceId;
private final Schema schema;
// .. lot of other fields here
}
现在我有以下代码,我在两个类之上迭代,并在给定schema
的情况下提取clientId
。
public Optional<Schema> getSchema(final String clientId) {
for (ProcessMetadata metadata1 : processMetadataList) {
if (metadata1.getClientId().equalsIgnoreCase(clientId)) {
String deviceId = metadata1.getDeviceId();
for (ProcMetadata metadata2 : procMetadataList) {
if (metadata2.getDeviceId().equalsIgnoreCase(deviceId)) {
return Optional.of(metadata2.getSchema());
}
}
}
}
return Optional.absent();
}
有没有更好的方法来获得我需要的东西,通过迭代上面的两个类而不是我的东西?我使用的是Java 7。
答案 0 :(得分:6)
您正在进行quadratic *搜索操作,这是不合适的。您可以通过首先从每个列表的id-&gt;对象创建mapping(在线性时间内)来在恒定时间内执行此操作。这看起来像这样:
// do this once, in the constructor or wherever you create these lists
// even better discard the lists and use the mappings everywhere
Map<String, ProcessMetadata> processMetadataByClientId = new HashMap<>();
for (ProcessMetadata process : processMetadataList) {
processMetadataByClientId.put(process.getClientId(), process);
}
Map<String, ProcMetadata> procMetadataByDeviceId = new HashMap<>();
for (ProcMetadata metadata2 : procMetadataList) {
procMetadataByDeviceId.put(proc.getDeviceId(), proc);
}
然后你的查询就变成了:
public Optional<Schema> getSchema(String clientId) {
ProcessMetadata process = processMetadataByClientId.get(clientId);
if (process != null) {
ProcMetadata proc = procMetadataByDeviceId.get(process.getDeviceId());
if (proc != null) {
return Optional.of(proc.getSchema());
}
}
return Optional.absent();
}
在Java 8中,您可以这样写:
public Optional<Schema> getSchema(String clientId) {
return Optional.fromNullable(processMetadataByClientId.get(clientId))
.map(p -> procMetadataByDeviceId.get(p.getDeviceId()))
.map(p -> p.getSchema());
}
*实际上,假设客户端ID是唯一的,您的算法是线性的,但它在技术上仍然是O(n ^ 2),因为您可能会触摸进程列表中每个元素的proc列表的每个元素。对算法稍作调整可以保证线性时间(再次假设唯一ID):
public Optional<Schema> getSchema(final String clientId) {
for (ProcessMetadata metadata1 : processMetadataList) {
if (metadata1.getClientId().equalsIgnoreCase(clientId)) {
String deviceId = metadata1.getDeviceId();
for (ProcMetadata metadata2 : procMetadataList) {
if (metadata2.getDeviceId().equalsIgnoreCase(deviceId)) {
return Optional.of(metadata2.getSchema());
}
}
// adding a break here ensures the search doesn't become quadratic
break;
}
}
return Optional.absent();
}
虽然使用地图当然可以确保恒定时间,这要好得多。
答案 1 :(得分:1)
我想知道可以用番石榴做些什么,并且偶然写下了这个热点。
import static com.google.common.collect.Iterables.tryFind
public Optional<Schema> getSchema(final String clientId) {
Optional<String> deviceId = findDeviceIdByClientId(clientId);
return deviceId.isPresent() ? findSchemaByDeviceId(deviceId.get()) : Optional.absent();
}
public Optional<String> findDeviceIdByClientId(String clientId) {
return tryFind(processMetadataList, new ClientIdPredicate(clientId))
.transform(new Function<ProcessMetadata, String>() {
String apply(ProcessMetadata processMetadata) {
return processMetadata.getDeviceId();
}
});
}
public Optional<Schema> findSchemaByDeviceId(String deviceId) {
return tryFind(procMetadataList, new DeviceIdPredicate(deviceId.get())
.transform(new Function<ProcMetadata, Schema>() {
Schema apply(ProcMetadata procMetadata) {
return processMetadata.getSchema();
}
});
}
class DeviceIdPredicate implements Predicate<ProcMetadata> {
private String deviceId;
public DeviceIdPredicate(String deviceId) {
this.deviceId = deviceId;
}
@Override
public boolean apply(ProcMetadata metadata2) {
return metadata2.getDeviceId().equalsIgnoreCase(deviceId)
}
}
class ClientIdPredicate implements Predicate<ProcessMetadata> {
private String clientId;
public ClientIdPredicate(String clientId) {
this.clientId = clientId;
}
@Override
public boolean apply(ProcessMetadata metadata1) {
return metadata1.getClientId().equalsIgnoreCase(clientId);
}
}
对不起。