String object cannot store whether the text is bold, italic. It has only a character array. Bold or Italic means a visual effect. You can try something like following to detect what are the text you should add effects. String str = “Google Plus style #ListViews are all the rage these days on #Android because … Read more
Scanner fkc = new Scanner (System.in); System.out.println(“Enter an Element or something: “); String unknown = fkc.nextLine(); char [] arr = unknown.toCharArray(); int lastLetterIndex = arr.length – 1; if(arr[lastLetterIndex]==’h’) { System.out.println(“ends with H”); } else { System.out.println(“ends with no H”); } 3 solved I want to get the last letter in a string and print true … Read more
These codes are not equal: C version allocates byte array where as Java’s – integer array C version does not do clamping of array elements as Java does So in Java you should make a byte array instead: byte[] tape = new byte[30000]; And remove clamping in adjust() function: // if(tape[dp] < 0) tape[dp] += … Read more
You are initialising your birthdaypeople array everytime in each iteration which is why the else block is never executed because for every j the birthdaypeople array is new and have false value. Take out the array declaration out of the for loop. boolean[] birthdaypeople = new boolean[n]; for (int j = 0; j < n; … Read more
What you described is known as Levenshtein distance. There is library called Apache commons-text that you can use to do it for you. int ops = new LevenshteinDistance().apply(string1, string2) Here is source code 3 solved Calculate number of Updation, Insertion and Deletion on a One String to reach Another String [closed]
Modify the body of the main method like this: public static void main(String[] arg) { Address address = new Address(“The Game”, “Seventh street”, “Generic Town”, “State”, “1234”); System.out.println(address.toString()); } 1 solved Pls help me, how can I make the code printout the full details of the address as an output instead of empty output [closed]
If different identifiers can have same values, your third map will keep the last parsed one. E.g. : File 1 : localId1 => “aaaa” localId2 => “bbbb” localId3 => “cccc” localId4 => “aaaa” File 2: localId1 => “1111” localId2 => “2222” localId3 => “3333” localId4 => “4444” Your first and second maps will store this … Read more
With the help of Scanner class you can take input from cmd and store each line into Array of String or In ArrayList collection. public static void main(String[] args) { ArrayList<String> in = new ArrayList<String>(); Scanner s = new Scanner(System.in); while (s.hasNextLine()) { String line = s.nextLine(); in.add(line); if (line != null && line.equalsIgnoreCase(“END”)) { … Read more
You just need to build the squares until you exceed the value N. Try: public static void main(String[] args) { Scanner entrada = new Scanner(System.in); int n = entrada.nextInt(); for (int i = 1; i * i <= n; i++) { System.out.println(i * i); } } 3 solved Print all the squares from natural numbers … Read more
Just use a regular for loop to iterate over all arrays at once: String[] item ={“[1]hotdog”, “[2]eggpie”,”[3]menudo”,”[4]pizza”,”[5]lumpia”}; int[] cost = {5, 10, 15, 20, 25}; int[] selling = {10,15,20,25,30,}; int[] qty = {2,4,6,8,10}; for (int i = 0; i < item.length; i++) { System.out.println(” ” +item[i]+”\t”+cost[i]+”\t\t”+selling[i]+”\t\t”+qty[i]); } I would recommend putting all of these fields … Read more
Line 46 is; actionBar.setNavigationMode(ActionBar.NAVIGATION_MODE_LIST); The only thing that can be null on here is actionBar, therefore getActionBar() is returning null, the reasons for this; getActionBar() returns null – Title bar isn’t visible. getActionBar returns null – You must request an action bar using getWindow().requestFeature(Window.FEATURE_ACTION_BAR);, you may think your info bar is an action bar. etc. … Read more
website: not really. You need msoffice to run the sheet, so unless you manually execute the sheet previously so that the data is available for read (not execute) you can’t do it with ms excel. You can do it with a programming language, depending on what you need and what you are familiar with you … Read more
The string you have mentioned is a JSON string. JSON (JavaScript Object Notation) is a lightweight data-interchange format. It is easy for humans to read and write. It is easy for machines to parse and generate. It is based on a subset of the JavaScript Programming Language, Standard ECMA-262 3rd Edition – December 1999. JSON … Read more
Try this code, I only change the position after part end of z as first and second reachable statement in on create function. @Override protected void onCreate(Bundle savedInstanceState) { //END OF PART Z super.onCreate(savedInstanceState); setContentView(R.layout.activity_main); //PART Z String consumerKey = “YII”; String consumerSecret = “YII”; String accessToken = “YII”; String accessTokenSecret = “YII”; //Instantiate a … Read more
In the Room class, the block variable (or one of its elements) is null. Because the error happens on this line: public void draw(Graphics g) { > for(int y=0;y<block.length;y++) { for(int x=0;x<block[0].length;x++) { block[y][x].draw(g); } } } I’d say you haven’t called define() (which seems to initialize the block variable) on your room by the … Read more