try boolean containsWord(String s, String w) { List<Character> list = new LinkedList<Character>(); for (char c : s.toCharArray()) { list.add(c); } for (Character c : w.toCharArray()) { if (!list.remove(c)) { return false; } } return true; } 0 solved Find word in random string
Revised Because strings are immutable their references aren’t copied. To get around this you can use a StringBuffer, but because you want to update a string and have it reflect on another string, your going to need to do some encapsulation of the StringBuffer’s to accomplish it. class StringChain { private StringBuffer[] buffers; public StringChain(StringBuffer[] … Read more
This will not visit all indexes of the array because array indexes begin at 0, not 1. In other words, the first element of the 2D array would be adjacencyMatrix[0][0] so you should start both of your iterations from 0. If the array has a length of 5, the largest index would therefore be four … Read more
The velocityThreashold variable in your example is not final, nor is it an instance variable and therefore cannot technically be overridden. What you can do is set the value of velocityThreashold to any value you want since it’s public. I think what you’re going to want to do is something like the following: public static … Read more
The problem has nothing to do with gmail integration. The problem is with eclipse “Unable to execute dex: Java heap space” and “Conversion to Dalvik format failed: Unable to execute dex: Java heap space” This comes when the eclipse has memory related issues. Try restarting eclipse and/or your system. It should work. Also you might … Read more
Yes, you are correct. By calling new you are creating a reference to an object, you are doing a similar operation when you declare a new String. 5 solved What is the error in each Java Statement?
You still haven’t asked any question, but I assume that your program gives wrong answer (as it really do) and you don’t know why. For example: for n=16 it prints: 1234 1234 1234 1234 The following code fragment is responsible: array[i][j]=j+1; System.out.print(array[i][j]); Firstly, you don’t print spaces, so numbers aren’t separated. Secondly, you have to … Read more
You can easily order by station without understanding the SQL query statements using SQLiteDatabase.query() method Cursor cursor = db.query(“TABLE_NAME”,new String[]{“COLUMN1″,”COLUMN2″,”COLUMN3″},null,null,null,null,”COLUMN1 ASC”); 0 solved Listview Order By Name
Note: if you want to really do it right, you can implement a subclass of JTextField that doesn’t even let the user enter characters that are invalid floating-point numbers: public class FloatField extends JTextField { public FloatField(int cols) { super(cols) } protected Document createDefaultModel() { return new FloatDocument(); } static class FloatDocument extends PlainDocument { … Read more
I have tested this in Java and prints the message: class Foo { void bar() { class Baz { void hi() { System.out.println(“Hi”); } } Baz baz = new Baz(); baz.hi(); } } For non-Java programmers, this would result very odd, but is the base for anonymous classes now I need to call methodTwo from … Read more
public static int lastDigitsSum(int total) { try (Scanner scan = new Scanner(System.in)) { String str = scan.next(); int count = 0; for (int i = str.length() – 1, j = 0; i >= 0 && j < total; i–, j++) { if (Character.isDigit(str.charAt(i))) count += str.charAt(i) – ‘0’; else throw new RuntimeException(“Input is not a … Read more
you can try this public static void main(String[] args) { String[] strings = {“one”, “one”, “one”, “one”, “one”, “one”, “one”, “one”, “one”, “one”, “one”, “one” , “two”, “two”, “two”, “two”, “two”, “two”, “two”, “two”, “two”, “two”, “two”, “two”, “two”, “two”, “two” , “three”, “three”, “three”, “three”, “three”, “three”, “three”, “three”, “three”, “three”, “three” , “four”, … Read more
You have switched the logic between resume and pause. Pause should persist your number. Resume should read the old number. 4 solved Anyone know whats wrong with my code? [closed]
Declare an array and then store the match results to that. ([^\”]*) captures any character but not of ” zero or more times. () called capturing group which is used to capture the characters which are matched by the pattern present inside that group. Later we could refer those captured characters through back-referencing. String s … Read more
You below lib :- https://github.com/chrisbanes/PhotoView XML <uk.co.senab.photoview.PhotoView android:id=”@+id/iv_photo” android:layout_width=”fill_parent” android:layout_height=”fill_parent” /> JAVA ImageView mImageView = (ImageView) findViewById(R.id.iv_photo); hope above lib will helps you. solved How to implement both pinch zoom and pan in android? [closed]