This is a chunk of code in Java, I'm trying to output random numbers from the tasks array, and to make sure none of the outputs are repeated, I put them through some other loops (say you have the sixth, randomly-chosen task "task[5]"; it goes through the for loop that will check it against every "tCheck" element, and while task[5] equals one of the tCheck elements, it will keep trying to find another option before going back to the start of the checking forloop... The tCheck[i] elements are changed at the end of each overall loop of output to the new random number settled on for the task element).
THE PROBLEM is that, despite supposedly checking each new random task against all tCheck elements, sometimes (not always) there are repeated tasks output (meaning, instead of putting out say 2,3,6,1,8,7,5,4, it will output something like 2,3,2,1,8,7,5,4, where "2" is repeated... NOT always in the same place, meaning it can sometimes end up like this, too, where "4" is repeated: 3,1,4,5,4,6,7,8)
int num = console.nextInt();
String[] tasks = {"1","2","3","4","5","6","7","8"};
String[] tCheck = {"","","","","","","",""};
for(int i = 0; i<= (num-1); i++){
int tNum = rand.nextInt(8);
for(int j = 0; j <=7; j++){
if(tasks[tNum].equals(tCheck[j])){
while(tasks[tNum].equals(tCheck[j])){
tNum = rand.nextInt(8);
}
j = 0;
}
}
tCheck[i] = tasks[tNum];
System.out.println(tasks[tNum]+" & "+tCheck[i]);
}
None of the other chunks of code affect this part (other than setting up Random int's, Scanners, so on; those are all done correctly). I just want it to print out each number randomly and only once. to never have any repeats. How do I make it do that?
Thanks in advance.
Firstly, don't use arrays. Use collections - they are way more programmer friendly.
Secondly, use the JDK's API to implement this idea:
randomise the order of your elements
then iterate over them linearly
In code:
List<String> tasks = Arrays.asList("1","2","3","4","5","6","7","8");
Collections.shuffle(tasks);
tasks.forEach(System.out::println);
Job done.
you can check if a certain value is inside your array with this approach.
for(int i = 0; i<= (num-1); i++){
int tNum = rand.nextInt(8);
boolean exist = Arrays.asList(tasks).contains(tNum);
while(!exist){
//your code
int tNum = rand.nextInt(8);
exist = Arrays.asList(tasks).contains(tNum);
}
}
if you are using an arraylist then you can check it with contains method since you are using an array we have to get the list from the array using asList() and then use the contains method. with the help of the while loop it will keep generating random numbers untill it generates a non duplicate value.
I used to created something similar using an ArrayList
public class Main {
public static void main(String[] args) {
String[] array = { "a", "b", "c", "d", "e" };
List<String> l = new ArrayList<String>(Arrays.asList(array));
Random r = new Random();
while(!l.isEmpty()){
String s = l.remove(r.nextInt(l.size()));
System.out.println(s);
}
}
}
I remove a random position in the list until it's empty. I don't use any check of content. I believe that is kind of effective (Even if I create a list)
Related
I have a small problem, I want to go through a list and compare two objects of the array. Each object has 3 elements, I use a StringTokenizer to be able to remove the separator, so each object has 3 elements. I would like to know how to make a method that gets the third element of each object and compare them. And if that element is less than another delete that element and the 2 before it.
I tried to make them with an iterator but I wouldn't know very well that it started from the 3 element and increased the position by 3.
Iterator<Integer> it = lisM.iterator();
int num;
while (it.hasNext()){
num = it.next();
System.out.println(num);
}
Is --> if, I was wrong to put it in the picture
This only answers part of your question. I could not understand the question completely, please edit it and I can edit my answer.
You should not remove items from a list whilst in a for loop, therefore you can, for example, create another boolean list with the same size divided by 3 and just fill it with true Booleans then set the position divided by 3 to false if you want to delete the three items. Then you can create a new list, iterate over the boolean list and add 3 "Objects" which are actually Strings (thanks #JB Nizet) at a time, every time the boolean list element is true. When it is false you just don't add the elements and by doing so you are practically deleting the two elements before that element together with that element.
You casted a String to an int, that does not work you have to parse the Strings.
I corrected some of your code and added the boolean list here:
ArrayList<String> lisM = new ArrayList<>(); // here I initialise the list as an array list with strings.
ArrayList<Boolean> booleanList = new ArrayList<>();
for (int i = 0; i < lisM.size() / 3; i++) {
booleanList.add(true);
}
for(int i = 3; i < lisM.size();i+=3) {
int m = Integer.parseInt(lisM.get(i)); // here I changed the casting to parsing and moved it out of the for loop, there is no need to initialize it again every single time since you do not change it in the second for loop.
for (int j = 6; j < lisM.size(); j += 6) {
int m1 = Integer.parseInt(lisM.get(j));// here I changed the casting to parsing again.
if (m > m1) { // this makes no sense here because you are going over all of the elements of the list and comparing them to all of them. But I kept it here for the sake of example.
booleanList.set(i/3,false);
}
// if you want to go over the whole list you will have to clear the list and start over again for every element.
}
}
and here is how you could create the new list without the elements you do not want:
ArrayList<String> newLisM = new ArrayList<>();
for (int i = 0; i <booleanList.size(); i++) {
if(booleanList.get(i))
for (int j = 0; j < 3; j++) {
newLisM.add(lisM.get(i+j));
}
}
This is a peice of my code, i am making a grid of 5x5 with random colors set to each section. I need to set the specified y_loc and x_loc in the list to the color randomly picked except i have not been able to find out how. It should be the second last line that is not operating as id like. I understand that i could do this in much much longer code but it would be nice to do it in less.
//making the map
ArrayList<ArrayList<String>> fullmap = new ArrayList<ArrayList<String>>();
ArrayList<String> y_row_0 = new ArrayList<String>();
ArrayList<String> y_row_1 = new ArrayList<String>();
ArrayList<String> y_row_2 = new ArrayList<String>();
ArrayList<String> y_row_3 = new ArrayList<String>();
ArrayList<String> y_row_4 = new ArrayList<String>();
//adding each row
fullmap.add(y_row_0);
fullmap.add(y_row_1);
fullmap.add(y_row_2);
fullmap.add(y_row_3);
fullmap.add(y_row_4);
Random rn = new Random();
//loop to randomly pick colors then set them to their destined locations
for (int y_loc = 0; y_loc < 6; y_loc++){
for (int x_loc = 0; x_loc < 6; x_loc++){
colorPicked = false;
while (!colorPicked){
int ranNum = rn.nextInt();
if (ranNum ==0){
if (redTot < 5) {
redTot += 1;
fullmap.set(y_loc).set(x_loc, "Red"));
colorPicked = true;
Since you have lists in list here, to set something at a specific location, you'll have to get the inner list and then perform the set on it.
The following should work:
fullmap.get(y_loc).set(x_loc, "Red"));
Also, since you seem to always have a 5x5 matrix, I'd recommend using a double array instead. That'd make that line:
fullmap[x_loc][y_loc] = "Red";
You should have something like this:
fullmap.get(y_loc).set(x_loc, "Red"));
Notice the "get". You are "getting" the list at the y location, which returns an array list, then calling "set" against that array list to set the actual value in the index of the "x_loc" value.
You need to make a couple of changes:
While declaring the sub lists, you need to make sure they have 5 empty/null elements. Otherwise set will throw IndexOutOfBoundsException. E.g. you need to declare the lists like this:
ArrayList<String> y_row_0 = Arrays.asList(new String[5]);//Assuming it will have 5 elements
While setting the element, you first need to get the corresponding sub list, e.g. the following needs to be changed from:
fullmap.set(y_loc).set(x_loc, "Red"));
to
fullmap.get(y_loc).set(x_loc, "Red"));
Others have already discussed the indexing issue. Apart from that, I believe that your conditionals may not be executing as you expect. nextInt() will return a reasonable uniform random number in the range of -2147483648 to 2147483647. You have a 1/2^64 chance of getting a 0. Reduce the random number range to something more reasonable. For example, nextInt(10) will return a random number between 0 and 9.
Furthermore, if the probability is too low, you will not get 5 reds all the time. To guarantee 5 picks and for the sake of computational efficiency, it is better to randomly pick array indices and evaluate whether a color is set or not, such as the following pseudo-code
int redTot = 0;
while ( redTot < 5 ) {
int r = rn.nextInt( 5 );
int c = rn.nextInt( 5 );
String color = fullmap.get( r ).get( c );
if ( color == null ) {
fullmap.get( r ).set( c, "Red" );
redTot++;
}
}
I'm currently working on a homework assignment for a beginner-level class and I need help building a program that tests if a sodoku solution presented as an int[][] is valid. I do this by creating helper methods that check both rows, columns and grids.
To check the column I call a method called getColumn that returns a column[]. When I test it out it works fine. I then pass it out on a method called uniqueEntries that makes sure that there are no duplicates.
Problem is, when I call my getColumn method, it returns an array consisting of only one number (for example 11111111, 22222222, 33333333). I have no idea why it does that. Here is my code:
int[][] sodokuColumns = new int[length][length];
for(int k = 0 ; k < sodokuPuzzle.length ; k++) {
sodokuColumns[k] = getColumn(sodokuPuzzle, k);
}
for (int l = 0; l < sodokuPuzzle.length; l++) {
if(uniqueEntries(sodokuColumns[l]) == false) {
columnStatus = false;
}
}
my helper is as follows
public static int[] getColumn(int[][] intArray, int index) {
int[] column = new int[intArray.length];
for(int i = 0 ; i < intArray.length ; i++) {
column[i] = intArray[i][index];
}
return column;
}
Thanks !
You said:
when I call my getColumn method, it returns an array consisting of only one number (for example 11111111, 22222222, 33333333).
I don't see any issue with your getColumn method other than the fact it's not even needed because getColumn(sodokuPuzzle, k) is the same as sodokuPuzzle[k]. If you're going to conceptualize your 2D array in such a way that your first index is the column then for your purpose of checking uniqueness you only need to write a method to get rows.
The issue you're having would seem to be with another part of your code that you did not share. I suspect there's a bug in the logic that accepts user input and that it's populating the puzzle incorrectly.
Lastly a tip for checking uniqueness (if you're allowed to use it) would be to create a Set of some kind (e.g. HashSet) and add all of your items (in your case integers) to that set. If the set has the same size as your original array of items then the items are all unique, if the size differs there are duplicates.
Let's say I want to generate 20 random numbers on a 8 by 6 grid.(8 columns, 6 rows) . Based on the answer from here:Creating random numbers with no duplicates, I wrote my code like this:
Random randomNumGenerator = new Random();
Set<Integer[][]> generated = new LinkedHashSet<Integer[][]>();
while (generated.size() < 20) {
int randomRows = randomNumGenerator.nextInt(6);
int randomColumns = randomNumGenerator.nextInt(8);
generated.add(new Integer[][]{{randomRows,randomColumns}});
}
In reality what happens is the Set see Integer[][]{{5,5}}; and Integer[][]{{5,5}};as NOT duplicate.Why? Even tho my purpose is to get 20 non-duplicate pair of numbers, this does not work. How do I fix this?
The Set checks for duplicates using the equals method (and also the hashCode method) of its inner type, but the Integer[][]'s equals method compares the memory addresses and not the contents.
Why do you use a Set of Integer[][] if you just want to store pairs?
Unfortunately, in Java there is no Pair class, but if you do not want to create your own, you can use the Map.Entry for that.
Random randomNumGenerator = new Random();
Set<Map.Entry<Integer, Integer>> generated = new LinkedHashSet<>();
while (generated.size() < 20) {
int randomRows = randomNumGenerator.nextInt(6);
int randomColumns = randomNumGenerator.nextInt(8);
generated.add(new AbstractMap.SimpleEntry<>(randomRows,randomColumns));
}
System.out.println(generated);
Array equals is == in Java, so an array is only equal to itself. Normaly you use Arrays.equals(array1, array2) to compare them by content, but in this case, arrays are simply the wrong choice. You can either create a bean, as rafalopez79 suggested of use an array of Collections (List in your case), as a List will compare the content on equals, see the documentation. Choice is pretty much yours, a bean would probably be a bit cleaner.
How about this code. I ran it through the debugger, it works nicely and yes, the contains() method checks the value of the Integer, not the reference. You can change the range of the random number as needed, I used 5 to facilitate testing. Yes I know it's not very robust, as written this will be an endless loop (because of the limited range of 5) but it's a simple example to make the point.
UPDATE: Actually this has a bug in that it won't check for uniqueness across all the rows, but that's easily fixed as well. I just re-read the original question and looking at the original code I'm not sure I know what you want exactly. If you just want a grid with 48 unique Intergers arranged 8 by 6 this will do it, but there are several ways to do this.
final int rows = 6;
final int cols = 8;
Random randomGenerator = new Random();
ArrayList[] grid = new ArrayList[rows];
for(int i=0; i<rows; i++)
{
grid[i] = new ArrayList<Integer>();
for(int j=0; j<cols; j++)
{
for(;;)
{
Integer newInt = new Integer(randomGenerator.nextInt(5));
if(!grid[i].contains(newInt))
{
grid[i].add(newInt);
break;
}
}
}
}
I'm trying to write a simple game where an enemy chases the player on a grid. I'm using the simple algorithm for pathfinding from the Wikipedia page on pathfinding. This involves creating two lists with each list item containing 3 integers. Here's test code I'm trying out to build and display such a list.
When I run the following code, it prints out the same numbers for each array in the ArrayList. Why does it do this?
public class ListTest {
public static void main(String[] args) {
ArrayList<Integer[]> list = new ArrayList<Integer[]>();
Integer[] point = new Integer[3];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 3; j++) {
point[j] = (int)(Math.random() * 10);
}
//Doesn't this line add filled Integer[] point to the
//end of ArrayList list?
list.add(point);
//Added this line to confirm that Integer[] point is actually
//being filled with 3 random ints.
System.out.println(point[0] + "," + point[1] + "," + point[2]);
}
System.out.println();
//My current understanding is that this section should step through
//ArrayList list and retrieve each Integer[] point added above. It runs, but only
//the values of the last Integer[] point from above are displayed 10 times.
Iterator it = list.iterator();
while (it.hasNext()) {
point = (Integer[])it.next();
for (int i = 0; i < 3; i++) {
System.out.print(point[i] + ",");
}
System.out.println();
}
}
}
First of all, several of the other answers are misleading and/or incorrect. Note that an array is an object. So you can use them as elements in a list, no matter whether the arrays themselves contain primitive types or object references.
Next, declaring a variable as List<int[]> list is preferred over declaring it as ArrayList<int[]>. This allows you to easily change the List to a LinkedList or some other implementation without breaking the rest of your code because it is guaranteed to use only methods available in the List interface. For more information, you should research "programming to the interface."
Now to answer your real question, which was only added as a comment. Let's look at a few lines of your code:
Integer[] point = new Integer[3];
This line creates an array of Integers, obviously.
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 3; j++) {
point[j] = (int)(Math.random() * 10);
}
//Doesn't this line add filled Integer[] point to the
//end of ArrayList list?
list.add(point);
//...
}
Here you assign values to the elements of the array and then add a reference to the array to your List. Each time the loop iterates, you assign new values to the same array and add another reference to the same array to the List. This means that the List has 10 references to the same array which has been repeatedly written over.
Iterator it = list.iterator();
while (it.hasNext()) {
point = (Integer[])it.next();
for (int i = 0; i < 3; i++) {
System.out.print(point[i] + ",");
}
System.out.println();
}
}
Now this loop prints out the same array 10 times. The values in the array are the last ones set at the end of the previous loop.
To fix the problem, you simply need to be sure to create 10 different arrays.
One last issue: If you declare it as Iterator<Integer[]> it (or Iterator<int[]> it), you do not need to cast the return value of it.next(). In fact this is preferred because it is type-safe.
Finally, I want to ask what the ints in each array represent? You might want to revisit your program design and create a class that holds these three ints, either as an array or as three member variables.
I would highly recommend to enclose the integer array of 3 numbers into a meaningful class, that would hold, display and control an array of 3 integers.
Then in your main, you can have an growing ArrayList of objects of that class.
You have an extra ) here:
element = (int[])it.next()); //with the extra parenthesis the code will not compile
should be:
element = (int[])it.next();
Besides the problem in the other answer, you cal it.next() two times, that cause the iterator move forward two times, obviously that's not what you want. The code like this:
element = (int[])it.next());
String el = (String)element;
But actually, I don't see you used el. Although it's legal, it seems meaningless.