Im not going to implement the solution for you but I’ll give you the pusedocode. for each letter1 in String1 for each letter2 in String2 if letter1 is letter2 then print its position So you know you will need a nested for loop and a comparison, it becomes a bit more complicated if you want … Read more
Since you need to adapt to different mechanics\protocols, you can implement Adapter pattern. Also, adapter can be chosen at runtime, you can also implement Factory pattern to instantiate an adapter. And then Strategy pattern to have adapters and factories. All this being done with IoC to inject dependencies like adapters or factories solved Abstract Factory … Read more
&& is a logical and operator || is the logical or operator Using De Morgan, the following: while(!gender.equals(MALE) && !gender.equals(FEMALE)) Can be translated to: while(!(gender.equals(MALE) || gender.equals(FEMALE))) (note the additional parenthesis and the placement of the ! before them). Both the above mean that the gender is neither MALE or FEMALE. Your other code: while(!gender.equals(MALE) … Read more
What you have <activity android:name=”.ScanResults” /> And when using intent Intent intent = new Intent(MainActivity.this, ScanResult.class); ScanResults and ScanResult are not the same. If Activity is not found you should get ActivityNotFoundException. This exception is thrown when a call to startActivity(Intent) or one of its variants fails because an Activity can not be found to … Read more
Try with this: private static final int MAX = 2; public static void sortZeroOneTwo(int[] input) { int []values = new int[MAX+1]; for (int element: input) { values[element]++; } int j = 0; for (int i = 0 ; i < values.length; i++){ for (int k = 0; k < values[i]; k++){ input[j++] = i; } … Read more
Mail sending has nothing to do with selenium.You have to write your separate java function for sending mail.You can call that function as per your requirement.For example if you want to send mail after test case execution is finished then you can call that function in tearDown method of your test class. 1 solved How … Read more
The IOException is part of the interface. Errors usally occur when a file is not present, the disk is full, you are missing reading or writing privledges or you have network connectivity issues. Depending on the implementation it might as well throw no errors at all even though its still declared in the interface. 2 … Read more
You can use negative lookbehind: String regex = “(?<!GoodClass\\.)\\bNiceMethod\\b”; 1 solved Regular expression to fetch lines evaluating only one part of a string, java
Your error log shows you have SecurityException: Permission Denial…with revoked permission android.permission.CAMERA. This means you are targetting API level 23 and the user has revoked the CAMERA permission. You should add code to check and request permission and handle permission acceptance/denial. Read more about it here. 3 solved Android – Menu Item Crashing on Click … Read more
I got the solution… private void method(String txt) { Pattern pattern = Pattern.compile((“md5_id=[a-zA-Z_0-9]{5,}”)); Matcher m = pattern.matcher(txt); while (m.find()) { Log.d(“log”, m.group()); } } 1 solved get matches words by regex java
Assuming that the .txt file contains one word per line, here is a simple solution: string[] keywords = System.IO.File.ReadAllLines(@”C:\Temp\keywords.txt”); List<string> nomatch = new List<string>(); System.Net.WebClient wc = new System.Net.WebClient(); foreach (string word in keywords) { string response = wc.DownloadString(“http://www.website.com/search.php?word=” + word); if (response != null && response.Contains(“No matches found”)) nomatch.Add(word); } if (nomatch.Count > 0) … Read more
Scanner in = new Scanner(new File(“path_to_file.txt”)); String var = in.next(); var = var.trim(); in.close(); Not the most efficient way, but one of the simplest. 0 solved How can I read a single word from a txt file and make it a variable in my code? [closed]
“Which is better” is always a matter of opinion. In your particular application, is execution time or memory use more of a problem? Then optimize for that aspect. In terms of leetcode, I understand it’s a sort of quiz, but for numbers that small, my response is “who cares?”. I don’t care about an execution-time … Read more
When you’re using Scanner.nextInt() the input is expected to be an integer. Inputting anything else, including a real number, will throw an InputMismatchException. To make sure invalid input doesn’t stop your program, use a try/catch to handle the exception: int num; try { num = sc.nextInt(); // Continue doing things with num } catch (InputMismatchException … Read more
Here is a test application to show you the trace of the program’s execution (requires Java 11 due to the use of String#repeat(int)): public class Main { private static int depth; public static void main(String[] args) { foo(0); } public static void foo(int n) { printDebug(“Invoked foo(” + n + “)”); if (n < 2) … Read more