I try to write a old maid.
After dealing cards,and sorting, i have two parts of card,one is playerDeck, one is computerDeck. now the pairs need to be removed.but i was stuck at this stage.
for example(just an example )
playerDeck:
'A♡', 'A♢', '8♡', '8♢', '8♠', 'Q♠', '2♠', '4♣', '7♢', '7♣', 'K♣', 'A♡', 'J♡', '9♣', '3♢'
computerDeck:
'3♡','3♣', '10♡','10♠','10♣', '6♡', 'K♡','K♢', 'A♣', 'A♠', '4♢', '7♡','7♠'
String q;
String p;
ArrayStringsTools AA=new ArrayStringsTools();//this is a class that i will use for removing item
for(int i=0;i<playerDeck.length-1;i++){
q=playerDeck[i];
q=q.substring(0,1);//i try to find the first character
p=playerDeck[i+1];//after finding first character, i can compare them,and if they are same, then i can remove them
p=p.substring(0,1);
if(q==p){
AA.removeItemByIndex(playerDeck,26,i);//this is the method that i used for removing same item,i will put this code below
AA.removeItemByIndex(playerDeck,26,i+1);//there are 51 cards in total,player has 26, computer has 25
}
}
public static int removeItemByIndex(String[] arrayOfStrings, int currentSize, int itemToRemove){//this is the method i used for removing item(first is the array of Deck, second is the size of Deck,third is the index of item to remove)
if( arrayOfStrings == null || currentSize > arrayOfStrings.length) {
System.out.println("ArrayStringsTools.removeItemByIndex: wrong call");
return currentSize;
}
if( itemToRemove < 0 || itemToRemove >= currentSize ) {
System.out.println("ArrayStringsTools.removeItem: item "
+ itemToRemove + " out of bounds. Array Unchanged.");
return currentSize;
}
int i;
for( i = itemToRemove; i < currentSize-1; i++){
arrayOfStrings[i] = arrayOfStrings[i+1];
}
arrayOfStrings[i]= null;
return currentSize-1;
i think i wrote correctly, but it doesnt show any difference compared with the origin.
the result should be:
playerDeck: '8♠', 'Q♠', '2♠', '4♣', 'K♣', 'A♡', 'J♡', '9♣', '3♢'
computerDeck:'10♣', '6♡', '4♢'
or is there another way to do this,because when a pair removed,there are two empty spaces, so... I've been struggling for a long time......
To compare the 1st character, change this line
if (q == p) {
to
if (q.charAt(0) == p.charAt(0)) {
Notice that q == p checks to see if the q and p refer to the same string, and do not look at the contents at all. If you want to compare full strings (or any other object that is not a char, an int, or so on) by content, you should use equals: q.equals(p) returns true only if both have the same content.
If you want to compare two strings,you can use 'equals',like
if(q.equals(p)){//q==p if true,they are save in the same location-this may not be what you want,and in this code it will be false forever.
}
Related
I need to solve a crossword given the initial grid and the words (words can be used more than once or not at all).
The initial grid looks like that:
++_+++
+____+
___+__
+_++_+
+____+
++_+++
Here is an example word list:
pain
nice
pal
id
The task is to fill the placeholders (horizontal or vertical having length > 1) like that:
++p+++
+pain+
pal+id
+i++c+
+nice+
++d+++
Any correct solution is acceptable, and it's guaranteed that there's a solution.
In order to start to solve the problem, I store the grid in 2-dim. char array and I store the words by their length in the list of sets: List<Set<String>> words, so that e.g. the words of length 4 could be accessed by words.get(4)
Then I extract the location of all placeholders from the grid and add them to the list (stack) of placeholders:
class Placeholder {
int x, y; //coordinates
int l; // the length
boolean h; //horizontal or not
public Placeholder(int x, int y, int l, boolean h) {
this.x = x;
this.y = y;
this.l = l;
this.h = h;
}
}
The main part of the algorithm is the solve() method:
char[][] solve (char[][] c, Stack<Placeholder> placeholders) {
if (placeholders.isEmpty())
return c;
Placeholder pl = placeholders.pop();
for (String word : words.get(pl.l)) {
char[][] possibleC = fill(c, word, pl); // description below
if (possibleC != null) {
char[][] ret = solve(possibleC, placeholders);
if (ret != null)
return ret;
}
}
return null;
}
Function fill(c, word, pl) just returns a new crossword with the current word written on the current placeholder pl. If word is incompatible with pl, then function returns null.
char[][] fill (char[][] c, String word, Placeholder pl) {
if (pl.h) {
for (int i = pl.x; i < pl.x + pl.l; i++)
if (c[pl.y][i] != '_' && c[pl.y][i] != word.charAt(i - pl.x))
return null;
for (int i = pl.x; i < pl.x + pl.l; i++)
c[pl.y][i] = word.charAt(i - pl.x);
return c;
} else {
for (int i = pl.y; i < pl.y + pl.l; i++)
if (c[i][pl.x] != '_' && c[i][pl.x] != word.charAt(i - pl.y))
return null;
for (int i = pl.y; i < pl.y + pl.l; i++)
c[i][pl.x] = word.charAt(i - pl.y);
return c;
}
}
Here is the full code on Rextester.
The problem is that my backtracking algorithm doesn't work well. Let's say this is my initial grid:
++++++
+____+
++++_+
++++_+
++++_+
++++++
And this is the list of words:
pain
nice
My algorithm will put the word pain vertically, but then when realizing that it was a wrong choice it will backtrack, but by that time the initial grid will be already changed and the number of placeholders will be reduced. How do you think the algorithm can be fixed?
This can be solved in 2 ways:
Create a deep copy of the matrix at the start of fill, modify and return that (leaving the original intact).
Given that you already pass around the matrix, this wouldn't require any other changes.
This is simple but fairly inefficient as it requires copying the matrix every time you try to fill in a word.
Create an unfill method, which reverts the changes made in fill, to be called at the end of each for loop iteration.
for (String word : words.get(pl.l)) {
if (fill(c, word, pl)) {
...
unfill(c, word, pl);
}
}
Note: I changed fill a bit as per my note below.
Of course just trying to erase all letter may erase letters of other placed words. To fix this, we can keep a count of how many words each letter is a part of.
More specifically, have a int[][] counts (which will also need to be passed around or be otherwise accessible) and whenever you update c[x][y], also increment counts[x][y]. To revert a placement, decrease the count of each letter in that placement by 1 and only remove letters with a count of 0.
This is somewhat more complex, but much more efficient than the above approach.
In terms of code, you might put something like this in fill:
(in the first part, the second is similar)
for (int i = pl.x; i < pl.x + pl.l; i++)
counts[pl.y][i]++;
And unfill would look something like this: (again for just the first part)
for (int i = pl.x; i < pl.x + pl.l; i++)
counts[pl.y][i]--;
for (int i = pl.x; i < pl.x + pl.l; i++)
if (counts[pl.y][i] == 0)
c[pl.y][i] = '_';
// can also just use a single loop with "if (--counts[pl.y][i] == 0)"
Note that, if going for the second approach above, it might make more sense to simply have fill return a boolean (true if successful) and just pass c down to the recursive call of solve. unfill can return void, since it can't fail, unless you have a bug.
There is only a single array that you're passing around in your code, all you're doing is changing its name.
See also Is Java "pass-by-reference" or "pass-by-value"?
You identified it yourself:
it will backtrack, but by that time the initial grid will be already
changed
That grid should be a local matrix, not a global one. That way, when you back up with a return of null, the grid from the parent call is still intact, ready to try the next word in the for loop.
Your termination logic is correct: when you find a solution, immediately pass that grid back up the stack.
I need to create a method which checks each element in my array to see if it is true or false, each element holds several values such as mass, formula, area etc for one compound, and in total there are 30 compounds (so the array has 30 elements). I need an algorithm to ask if mass < 50 and area > 5 = true .
My properties class looks like:
public void addProperty (Properties pro )
{
if (listSize >=listlength)
{
listlength = 2 * listlength;
TheProperties [] newList = new TheProperties [listlength];
System.arraycopy (proList, 0, newList, 0, proList.length);
proList = newList;
}
//add new property object in the next position
proList[listSize] = pro;
listSize++;
}
public int getSize()
{
return listSize;
}
//returns properties at a paticular position in list numbered from 0
public TheProperties getProperties (int pos)
{
return proList[pos];
}
}
and after using my getters/setters from TheProperties I put all the information in the array using the following;
TheProperties tp = new properties();
string i = tp.getMass();
String y = tp.getArea();
//etc
theList.addProperty(tp);
I then used the following to save an output of the file;
StringBuilder builder = new StringBuilder();
for (int i=0; i<theList.getSize(); i++)
{
if(theList.getProperties(i).getFormatted() != null)
{
builder.append(theList.getProperties(i).getFormatted());
builder.append("\n");
}
}
SaveFile sf = new SaveFile(this, builder.toString());
I just cant work out how to interrogate each compound individually for whether they reach the value or not, reading a file in and having a value for each one which then gets saved has worked, and I can write an if statement for the requirements to check against, but how to actually check the elements for each compound match the requirements? I am trying to word this best I can, I am still working on my fairly poor java skills.
Not entirely sure what you are after, I found your description quite hard to understand, but if you want to see if the mass is less than 50 and the area is greater than 5, a simple if statement, like so, will do.
if (tp.getMass() < 50 && tp.getArea() > 5) {}
Although, you will again, have to instantiate tp and ensure it has been given its attributes through some sort of constructor.
Lots of ways to do this, which makes it hard to answer.
You could check at creation time, and just not even add the invalid ones to the list. That would mean you only have to loop once.
If you just want to save the output to the file, and not do anything else, I suggest you combine the reading and writing into one function.
Open up the read and the write file
while(read from file){
check value is ok
write to file
}
close both files
The advantage of doing it this way are:
You only loop through once, not three times, so it is faster
You never have to store the whole list in memory, so you can handle really large files, with thousands of elements.
In case the requirements changes, you can write method that uses Predicate<T>, which is a FunctionalInterface designed for such cases (functionalInterfaces was introduced in Java 8):
// check each element of the list by custom condition (predicate)
public static void checkProperties(TheList list, Predicate<TheProperties> criteria) {
for (int i=0; i < list.getSize(); i++) {
TheProperties tp = list.get(i);
if (!criteria.apply(tp)) {
throw new IllegalArgumentException(
"TheProperty at index " + i + " does not meet the specified criteria");
}
}
}
If you want to check if mass < 50 and area > 5, you would write:
checkProperties(theList, new Predicate<TheProperties> () {
#Override
public boolean apply(TheProperties tp) {
return tp.getMass() < 50 && tp.getArea() > 5;
}
}
This can be shortened by using lambda expression:
checkProperties(theList, (TheProperties tp) -> {
return tp.getMass() < 50 && tp.getArea() > 5;
});
Alright, I'm making a method that should be able to remove objects from an array list through the use of a string input.
Say I want to remove the following numbers: {1,2,4,3,3,1} from an arraylist. How can I ensure that it only removes 1 & 3 twice and 4 & 2 once?
What I have is:
mv.displayMessages("choosedicestokeep");
String in = mv.getInput();
for (char c : in.toCharArray()) {
int x = Character.getNumericValue(c);
for (Iterator<Integer> it = rollingHand.iterator(); it.hasNext(); ){
int i = it.next();
if (x == i) {
finalHand[finalArrIndex] = i;
it.remove();
finalArrIndex++;
}
}
}
But this checks the arraylist "RollingHand" and removes ALL instances of a number and not the number of times I write a number which is what I want.
So if i enter {1,1,1,2,2,4} it should remove three 1s, two 2s and one 4.
https://stackoverflow.com/users/4584292/mike Solved the obvious answer.
Breaking to a statement outside the inner loop solved the problem.
The method doesn't return anything because it sets a private int[] finalHand in the class which is later accessed by other methods.
All cred to Mike!
Folks, my method needs to add a new element into already a sorted list, i.e. in a proper position. The point is the method has to add the objects in a diagonal sort. For example,
board.set(1,1,11);
board.set(2,4,33);
board.set(3,4,66);
board.set(3,2,44);
board.set(3,3,55);
board.set(1,4,88);
board.set(0,2,77);
board.set(0,5,99);
board.set(2,1,22);
The result should be:
[(2,1,22), (3,2,44), (1,1,11), (3,3,55), (3,4,66), (0,2,77), (2,4,33), (1,4,88), (0,5,99)]
But my program prints this:
[(3,4,66), (3,3,55), (3,2,44), (2,4,33), (2,1,22), (1,4,88), (1,1,11), (0,5,99), (0,2,77)]
i.e. it does not put the objects into the proper positions.
I have a LinkedList<RowColElem<T>>leftDiagSeq where the objects are added and put into a proper position "on the go". What is my code missing?
NOTE: Im not allowed to use comparators, comparable interface!
Code
LinkedList<RowColElem<T>> rowColSeq;
private void sortedLeftDiagSeq(int row, int col, T x){
RowColElem<T> object = new RowColElem<T>(row, col, x);
ListIterator<RowColElem<T>> iter = leftDiagSeq.listIterator();
RowColElem<T> inListObject;
boolean added = false;
while(iter.hasNext()){
inListObject = iter.next();
if(object.getRow()-1 < inListObject.getRow() ||
object.getRow()-1 == inListObject.getRow() &&
object.getCol()-1 < inListObject.getCol()){
if( iter.hasPrevious() ){
iter.add(object);
}
}
}
}
Primary criterion is an elements "distance" from the main diagonal, negative distances indicating the lower triangle matrix.
if( object.getCol() - object.getRow() < inListObject.getCol() - inListObject.getRow()
||
object.getCol() - object.getRow() == inListObject.getCol() - inListObject.getRow() &&
object.getCol() < inListObject.getCol()){ ... }
I'm not sure about the last term. The data you've provided would also produce the expected result if row numbers are used to break the tie of equal distance from the main diagonal. Probably this doesn't matter since you want order from top-left to bottom-right within one diagonal.
I found recently the default renderable sort function in LibGDX wasn't quite up to my needs. (see; Draw order changes strangely as camera moves? )
Essentially a few objects rendered in front when they should render behind.
Fortunately, the renderables in question always have a guarantied relationship. The objects are attached to eachother so when one moves the other moves. One object can be seen as being literally "pinned" to the other, so always in front.
This gave me the idea that if I specified a "z-index" (int) and "groupname" (String) for each object, I could manually take over the draw order, and for things with the same groupname, ensure they are positioned next to eachother in the list, in the order specified by the z-index. (low to high)
//For example an array of renderables like
0."testgroup2",11
1."testgroup",20
2."testgroup2",10
3.(no zindex attribute)
4."testgroup",50
//Should sort to become
0."testgroup",20
1."testgroup",50
2.(no zindex attribute)
3."testgroup2",10
4."testgroup2",11
// assuming the object2 in testgroup2 are closer to the camera, the one without a index second closest, and the rest furthest<br>
//(It is assumed that things within the same group wont be drastically different distances)
I implemented a sort system in libgdx to do this as followed;
/**
* The goal of this sorter is to sort the renderables the same way LibGDX would do normally (in DefaultRenderableSorter)<br>
* except if they have a ZIndex Attribute.<br>
* A Zindex attribute provides a groupname string and a number.<br>
* Renderables with the attribute are placed next to others of the same group, with the order within the group determined by the number<br>
*
* For example an array of renderables like;<br><br>
* 0."testgroup",20<br>
* 1."testgroup2",10<br>
* 2.(no zindex attribute)<br>
* 3."testgroup",50<br>
* <br>Should become;<br><br>
* 0."testgroup",20<br>
* 1."testgroup",50<br>
* 2.(no zindex attribute)<br>
* 3."testgroup2",10<br>
* <br>
* assuming the object in testgroup2 is closer to the camera, the one without a index second closest, and the rest furthest<br>
* (It is assumed that things within the same group wont be drastically different distances)<br>
*
* #param camera - the camera in use to determine normal sort order when we cant place in a existing group
* #param resultList - an array of renderables to change the order of
*/
private void customSorter(Camera camera, Array<Renderable> resultList) {
//make a copy of the list to sort. (This is probably a bad start)
Array <Renderable> renderables = new Array <Renderable> (resultList);
//we work by clearing and rebuilding the Renderables array (probably not a good method)
resultList.clear();
//loop over the copy we made
for (Renderable o1 : renderables) {
//depending of if the Renderable as a ZIndexAttribute or not, we sort it differently
//if it has one we do the following....
if (o1.material.has(ZIndexAttribute.ID)){
//get the index and index group name of it.
int o1Index = ((ZIndexAttribute)o1.material.get(ZIndexAttribute.ID)).zIndex;
String o1GroupName = ((ZIndexAttribute)o1.material.get(ZIndexAttribute.ID)).group;
//setup some variables
boolean placementFound = false; //Determines if a placement was found for this renderable (this happens if it comes across another with the same groupname)
int defaultPosition = -1; //if it doesn't find another renderable with the same groupname, this will be its position in the list. Consider this the "natural" position based on distance from camera
//start looping over all objects so far in the results (urg, told you this was probably not a good method)
for (int i = 0; i < resultList.size; i++) {
//first get the renderable and its ZIndexAttribute (null if none found)
Renderable o2 = resultList.get(i);
ZIndexAttribute o2szindex = ((ZIndexAttribute)o2.material.get(ZIndexAttribute.ID));
if (o2szindex!=null){
//if the renderable we are comparing too has a zindex, then we get its information
int o2index = o2szindex.zIndex;
String o2groupname = o2szindex.group;
//if its in the same group as o1, then we start the processing of placing them nexto eachother
if (o2groupname.equals(o1GroupName)){
//we either place it in front or behind based on zindex
if (o1Index<o2index){
//if lower z-index then behind it
resultList.insert(i, o1);
placementFound = true;
break;
}
if (o1Index>o2index){
//if higher z-index then it should go in front UNLESS there is another of this group already there too
//in which case we just continue (which will cause this to fire again on the next renderable in the inner loop)
if (resultList.size>(i+1)){
Renderable o3 = resultList.get(i+1);
ZIndexAttribute o3szindex = ((ZIndexAttribute)o3.material.get(ZIndexAttribute.ID));
if (o3szindex!=null){
String o3groupname = o3szindex.group;
if (o3groupname!=null && o3groupname.equals(o1GroupName)){
//the next element is also a renderable with the same groupname, so we loop and test that one instead
continue;
}
}
}
// Gdx.app.log("zindex", "__..placeing at:"+(i+1));
//else we place after the current one
resultList.insert(i+1, o1);
placementFound = true;
break;
}
}
}
//if no matching groupname found we need to work out a default placement.
int placement = normalcompare(o1, o2); //normal compare is the compare function in DefaultRenderableSorter.
if (placement>0){
//after then we skip
//(we are waiting till we are either under something or at the end
} else {
//if placement is before, then we remember this position as the default (but keep looking as there still might be matching groupname, which should take priority)
defaultPosition = i;
//break; //break out the loop
}
}
//if we have checked all the renderables positioned in the results list, and none were found with matching groupname
//then we use the defaultposition to insert it
if (!placementFound){
//Gdx.app.log("zindex", "__no placement found using default which is:"+defaultPosition);
if (defaultPosition>-1){
resultList.insert(defaultPosition, o1);
} else {
resultList.add(o1);
}
}
continue;
}
//...(breath out)...
//ok NOW we do placement for things that have no got a ZIndexSpecified
boolean placementFound = false;
//again, loop over all the elements in results
for (int i = 0; i < resultList.size; i++) {
Renderable o2 = resultList.get(i);
//if not we compare by default to place before/after
int placement = normalcompare(o1, o2);
if (placement>0){
//after then we skip
//(we are waiting till we are either under something or at the end)
continue;
} else {
//before
resultList.insert(i, o1);
placementFound = true;
break; //break out the loop
}
}
//if no placement found we go at the end by default
if (!placementFound){
resultList.add(o1);
};
} //go back to check the next element in the incomeing list of renderables (that is, the copy we made at the start)
//done
}
//Copy of the default sorters compare function
//;
private Camera camera;
private final Vector3 tmpV1 = new Vector3();
private final Vector3 tmpV2 = new Vector3();
public int normalcompare (final Renderable o1, final Renderable o2) {
final boolean b1 = o1.material.has(BlendingAttribute.Type) && ((BlendingAttribute)o1.material.get(BlendingAttribute.Type)).blended;
final boolean b2 = o2.material.has(BlendingAttribute.Type) && ((BlendingAttribute)o2.material.get(BlendingAttribute.Type)).blended;
if (b1 != b2) return b1 ? 1 : -1;
// FIXME implement better sorting algorithm
// final boolean same = o1.shader == o2.shader && o1.mesh == o2.mesh && (o1.lights == null) == (o2.lights == null) &&
// o1.material.equals(o2.material);
o1.worldTransform.getTranslation(tmpV1);
o2.worldTransform.getTranslation(tmpV2);
final float dst = (int)(1000f * camera.position.dst2(tmpV1)) - (int)(1000f * camera.position.dst2(tmpV2));
final int result = dst < 0 ? -1 : (dst > 0 ? 1 : 0);
return b1 ? -result : result;
}
As far as I can tell my customSorter function produces the order I want - the renderables now look like they are drawn in the right order.
However, this also seems like a hackjob, and I am sure my sorting algorithm is horrendously inefficient.
I would like advice on how to either;
a) Improve my own algorithm, especially in regards to any quirks to bare in mind when doing cross-platform LibGDX development (ie, array types, memory management in regards to android/web etc)
b) Alternative more efficient solutions having a similar "z index override" of the normal draw-order sorting.
Notes;
. The grouping is necessary. This is because while things are firmly stuck relatively to eachother within a group, groups themselves can also move about in front/behind eachother. (but not between). This makes it tricky to do a "global" override of the draw order, rather then a local one per group.
. If it helps, I can add/change the zindexattribute object in any way.
. I am thinking somehow "pre-storeing" each group of objects in a array could help things, but not 100% sure how.
First of all do never copy a list if not needed. The list with renderables could be really huge since it also could contain resources. Copying will be very very slow. If you need something local and you need performance try to make it final since it can improve the performance.
So a simple approach would be the default sorting of Java. You need to implement a Comperator for your class for example the Class with z index could look like this:
public class MyRenderable {
private float z_index;
public MyRenderable(float i)
{
z_index = i;
}
public float getZ_index() {
return z_index;
}
public void setZ_index(float z_index) {
this.z_index = z_index;
}
}
If you want a faster sort since your list wont change that much on runtime you could implement a insertion sort since it does a faster job if the list is kind of presorted. If it is not pre sorted it does take longer but in general it should only be the first sort call where it is alot disordered in your case.
private void sortList(ArrayList<MyRenderable> array) {
// double starttime = System.nanoTime();
for (int i = 1; i < array.size(); i++) {
final MyRenderable temp = array.get(i);
int j = i - 1;
while (j >= 0 && array.get(j).getZ_index() < temp.getZ_index()) {
array.set(j + 1, array.get(j));
j--;
}
array.set(j + 1, temp);
}
// System.out.println("Time taken: " + (System.nanoTime() - starttime));
}
To use this method you simply call it with your Array
sortList(renderbales);
In your case you need to take care of the ones that do not have a Z index. Maybe you could give them a 0 since they'll get sorted at the right position(i guess). Else you can use the given methods in z case and the regular in no z case as you do already.
After the conversation in the comments. I dont think it is a good idea to push everything into one list. It's hard to sort and would be very slow. A better approach would be a list of groups. Since you want to have groups, programm a group. Do not use String names, use IDs or types (way more easy to sort and it doesn't really matter). So a simple group would be this:
public class Group{
//think about privates and getters or methods to add things which also checks some conditions and so on
public int groupType;
public ArrayList<MyRenderable> renderables;
}
And now all your groups into a list. (this contains all your renderbales then)
ArrayList<Group> allRenderables = new ArrayList<>();
Last but not least sort the groups and sort the renderables. Since i dont think that your group ids/names will change on runtime, sort them once or even use a SortedSet instead of a ArrayList. But basically the whole sorting looks like this:
for(Group g: allRenderables)
sortRenderables(g.renderables); //now every group is sorted
//now sort by group names
sortGroup(allRenderables);
With the following insertionsorts as shown above
public static void sortRenderables(ArrayList<MyRenderable> array) {
for (int i = 1; i < array.size(); i++) {
final MyRenderable temp = array.get(i);
int j = i - 1;
while (j >= 0 && array.get(j).getZ_index() < temp.getZ_index()) {
array.set(j + 1, array.get(j));
j--;
}
array.set(j + 1, temp);
}
}
public static void sortGroup(ArrayList<Group> array) {
for (int i = 1; i < array.size(); i++) {
final Group temp = array.get(i);
int j = i - 1;
while (j >= 0 && array.get(j).groupType < temp.groupType) {
array.set(j + 1, array.get(j));
j--;
}
array.set(j + 1, temp);
}
}