我们假设要创建一个程序，该程序对在《詹姆斯国王圣经》的文本文件中找到的所有唯一单词进行计数，并打印出最常用的单词。我计算唯一单词数量的功能太慢了：
public int parseBook(File fileName) throws FileNotFoundException {
scan = new Scanner(fileName);
String currentWord = scan.next();
Word firstWord = new Word(currentWord);
allWords.add(firstWord);
wordCount++;
while(scan.hasNext()) {
currentWord = scan.next();
Word newWord = new Word(currentWord);
if(!allWords.contains(newWord)) {
uniquewordList.add(newWord);
}
else {
newWord.incrementFrequency();
}
allWords.add(newWord);
}
return uniquewordList.size();
}
答案 0 :(得分：0)
这很简单。只需使用 public class Client extends AsyncTask<String, Integer, Boolean> {
Socket socket;
String hostAdd;
InputStream inputStream;
OutputStream outputStream;
public Client(InetAddress hostAddress) {
hostAdd = hostAddress.getHostAddress();
socket = new Socket();
}
@Override
protected Boolean doInBackground(String... strings) {
boolean result = false;
try {
socket.connect(new InetSocketAddress(hostAdd, 8888), 5000);
result = true;
return result;
} catch (IOException e) {
e.printStackTrace();
result = false;
return result;
}
}
public void writeData(final byte[] bytes) {
new Thread(new Runnable() {
@Override
public void run() {
try {
outputStream.write(bytes);
} catch (IOException e) {
e.printStackTrace();
}
}
}).start();
btnSend.setVisibility(View.VISIBLE);
}
@Override
protected void onPostExecute(Boolean result) {
if(result) {
try {
inputStream = socket.getInputStream();
outputStream = socket.getOutputStream();
} catch (IOException e) {
e.printStackTrace();
}
new Thread(new Runnable(){
public void run() {
byte[] buffer = new byte[1024];
int x;
while (socket!=null) {
try {
x = inputStream.read(buffer);
if(x>0) {
handler.obtainMessage(MESSAGE_READ,x,-1,buffer).sendToTarget();
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
}).start();
btnSend.setVisibility(View.VISIBLE);
} else {
Toast.makeText(getApplicationContext(),"could not create sockets",Toast.LENGTH_SHORT).show();
//restart socket assignment process
}
}
}
并输入所有单词即可。将 O（1）插入一个单词到Set中。
Set