Try this code,you may got the expected answer, String str=”a b c”; boolean a1 = true ; StringBuilder strbu = new StringBuilder(); for(int i=0 ; i<str.length();i++) { Character b = str.charAt(i); if(b.equals(‘ ‘) && a1 == true ) { a1 = false; strbu = strbu.append(“AA”); } else { strbu = strbu.append(b); } } Log.e(“str”,”string:”+strbu.toString()); solved … Read more
You are creating an anonymous object. After the execution of constructor; the GC will clean up this object. https://www.quora.com/What-is-anonymous-object-instantiation solved How this code Compile and Run?
Indeed, you can get the date from a Calendar object: Calendar cal = new GregorianCalendar() Date d; cal.add(Calendar.DATE, 3); d = cal.getTime(); solved How to compute calendar date?
What is a simple algorithm / approach to store key-value (number-number) pairs *in disk* for quick access by key? [closed] solved What is a simple algorithm / approach to store key-value (number-number) pairs *in disk* for quick access by key? [closed]
In Java, a static method is associated with the class and may be called without an instance. And therefore cannot access instance variables either. In c++, static keyword has several meanings depending on context. A static member function (which is what class methods are typically called in c++) is quite similar to static methods in … Read more
Exception classification help use understand what exactly is the issue at hand easily even before we see the description. It is a first sigth informer to the developer to understand the issue. For example if you are getting a NullPointerException / ArithemeticException and if you just say “Exception occured” can we understand anything? without a … Read more
Well, your question is not clear. But as far as I understand you need help to get the date. You can use Calendar() or Date(), use something like this Calendar.get(Calendar.MONTH)to get the month and year (or simply parse to string and substring where you want). Regarding the number at the beginning, I will assume (again, … Read more
nextDouble() leaves the newline in the input stream, so when you then call nextLine().charAt(0) the result of nextLine() is an empty string. Remove the newline from the stream, or use next(), as you did above. 5 solved Exception in thread “main” java.lang.StringIndexOutOfBoundsException: String index out of range: 0 [closed]
JLS 15.20.2 states: If a cast of the RelationalExpression to the ReferenceType would be rejected as a compile-time error (ยง15.16), then the instanceof relational expression likewise produces a compile-time error. In such a situation, the result of the instanceof expression could never be true. In this case (Intf) obj is not a compile time error … Read more
You can solve the problem by following the steps below like QBrute correctly pointed out. public String countMatches(String main) { //Create an array of the alphabets length main = main.toLowerCase(); int[] foundArray = new int[26]; String output = “”; //Create an alphabets array String[] alphabets = “a b c d e f g h i … Read more
Like @CommonsWare said, you need to handle your specific time logic. Why not use an AlarmManager which will call a Service in which your code for sending the sms actually “happens”. If you do use an AlarmManager, please note that different versions of Android will (likely) be handled by different methods related to AlarmManager, possibly … Read more
The most likely case here can be that the answer has leading/trailing white spaces. You need to trim() those out. if (answer.trim().equals(“OK”)) { You did not initially mention in the question that the answer had extra string besides “OK”. In that case, you either need to remove other stuffs(using a regex may be), or use … Read more
If the length of user[] and pass[] is always equal, you can try this: Update: This is much more simpler: String currentUser = “empty”; boolean login=false; for(int j = 0; j < user.length; j++){ if( username.equals(user[j]) && password.equals(pass[j])){ currentUser = user[j]; login=true; break; } } if(login) System.out.println(“Hello ” + currentUser+ “!” ); else System.out.println(“Incorrect Login!” … Read more
I guess it is wrong in your sense. I suspect you’re looking for this. String pattern = “(\\D*)(\\d+)(.*)”; Maybe you want to check reluctant quantifiers too. http://docs.oracle.com/javase/tutorial/essential/regex/quant.html 13 solved Is this regex pattern wrong? [closed]
for (int i = 1; i <= input; i++) { String line = “”; for (int k = 2; k >= 0; k–) { line += ((i >> k) & 1) == 1 ? “1” : “0”; } System.out.println(line); } That uses two for loops. solved Printing Binary values using nested loops [closed]