ｃｏｐｅ函数是把string对象字符串去掉空格，然后返回string对象
str是输入的字符串对象
//计算每个字母出现的次数
String str1=cope(str);
int n=1;
do{
//从字符串的第二个数开始比较，计算地一个字符出现的次数
for(int j=1;j<str1.length();j++){
if(str1.charAt(j)==str1.charAt(0))
n++;
}
//将统计过的字母全部转化为空格
str1=str1.replace(str1.charAt(0),' ');
//输出统计字母出现的总次数
System.out.println("字母"+str1.charAt(0)+"的个数为："+n);
//调用cope函数将str1字符串空格去掉
str1=cope(str1);
//n重置为1
n=1;
}while(str1.length()!=0);
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- Java code
没有可用分，求解决啊import java.io.*;import java.net.*;import java.awt.*;import java.awt.event.*;class liaot extends Frame{ Socket socket=null; private TextArea text1; private TextField text2;//图形界面 liaot(){ text1 = new TextArea(); text2 = new TextField(); add(text1,"Center"); add(text2,"South"); event(); setBounds(300,300,300,300); setVisible(true); }//事件响应 public void event(){ addWindowListener(new WindowAdapter(){ public void windowClosing(WindowEvent e){ System.exit(0); } }); text2.addActionListener(new ActionListener(){ public void actionPerformed(ActionEvent e){ try{ socket = new Socket("192.168.30.129",3334); }catch(Exception ee){} new Thread(new clientsend()).start(); new Thread(new clientrec()).start(); text2.setText(""); } }); } public static void main(String[] args)throws Exception { new liaot(); }//发送 class clientsend implements Runnable { public void run(){ byte[]buf = text2.getText().trim().getBytes(); String str = new String(buf,0,buf.length); System.out.println(str); text1.append(str); try{ OutputStream out = socket.getOutputStream(); out.write((str+"\r\n").getBytes()); out.flush(); }catch(Exception e){} } }//接收 class clientrec implements Runnable { public void run(){ byte[] buf = new byte[1024]; try{ InputStream in = socket.getInputStream(); in.read(buf); String str = new String(buf,0,buf.length); text1.append(str); }catch(Exception e){} } }
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用MAP做的：
- Java code
public static void count(String info) { Map<Character, Integer> map = new HashMap<Character, Integer>(); for (int i = 0; i < info.length(); i++) { Integer count = map.get(info.charAt(i)); if (count == null) { map.put(info.charAt(i), 1); } else { map.put(info.charAt(i), count + 1); } } for (Character c : map.keySet()) { System.out.println(c + " : " + map.get(c)); } }
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你代码没问题，除了那个cope，还有这样效率太低了。