I have a class that "bans" a player, but I don't want that player to be banned if his name is within a string array. I could loop through the length of the array and use booleans, but there has to be an easier way? I saw something that said just put if a condition is met, put return; and it'll stop all code running below that if statement.
Edit: Thanks for all the help! You were all helpful, even if you're one of the people that downvoted this, which is 4 people at least.
You could make a method that checks if the player is in the array of Strings and yes if you use return in a void method the method will just end.
For example
public void returnUsage(int n)
{
if(n==1)
{
return;
}
System.out.println("n doesn't equal 1.");
}
But it would probably be best to use an if and else to skip the code you don't want to run if the condition is not met. http://docs.oracle.com/javase/tutorial/java/nutsandbolts/if.html
An array is the wrong data structure for this task. Add the players to a Set<> allowedPlayers = new HashSet<String>() then use if(allowedPlayers.contains(name)) return; to conditionally exit the method based on whether the name of the player is in the set. The set will remain fast when the number of names gets large. With an array scanning the array will be slow when the array is big.
Related
*EDIT - SOLVED: After instantiating the Scanner Object, I used a delimiter as follows:
scanner.useDelimiter("");
Prior to this, I did try a delimiter that looked something like this (the exact code is available on Stack Overflow):
scanner.useDelimiter("\\p{javaWhitespace}");
...but it didn't work very well.
Thank you, everyone. If you're having this very same issue, try the first delimiter. If it doesn't work, upgrade your JDK to 13 then try it again.
Ok, my goal is to have a user input a credit card number which I would then like to store in an ArrayList of Integers and subsequently pass this list to my functions which will perform the Luhn algorithm in order to validate the provided number. Once the user presses Enter, the processing begins. This is a console application, nothing fancy.
Everything works beautifully...except the user-input part. None of the user-input is being stored into the declared ArrayList. I've inserted a print message to give me the size of the list just after the pertinent while-loop and....yep, 0. I also pass this list into a custom lengthChecker(ArrayList<Integer> list){} function subsequent to the relevant while-loop and it's printing my custom error-message.
I have declared local int variables within the scope of the while-loop and that wasn't helping much. I have tried getting the user's input as Strings and storing them in an ArrayList<String> list; then parsing the input but that didn't work very well (especially as I need the Enter key to behave as a delimiter such that the next steps can take place)
Anyways, here is the code to the function in question. Am I missing something obvious or should I just quit programming?
public void userInput() {
Scanner scanner = new Scanner(System.in);
List<Integer> list = new ArrayList<Integer>();
System.out.println("Please input the card-number to be checked then press Enter: ");
while(scanner.hasNextInt()) {
list.add(scanner.nextInt());
}
System.out.println("Length of list: " + list.size());
listLengthChecker(list);
scanner.close();
}
Thank you in advance.
I don't have the full context on all the code you've written to be able to solve your problem, but I can guess at what's going on. If you want to run any user I/O (such as the scanner), it must occur within the main method. I can only assume that you run your userInput() function within the main method in your class. However, because your userInput() function doesn't have the static keyword in its definition, it can't be accessed without initialising an object of the class - but as far as I can tell from your code, there is no object that the method could refer to. Add the static keyword (i.e. initialise the method as public static void userInput()) to be able to run the function as you intend.
As for the while loop - there's a small chance that this is a difference in Java versions (I use Java 11), but while(scanner.hasNextInt()) won't stop being true at the end of your line or when you press enter - only when you insert something (such as a character) that cannot be interpreted as an integer.
This while loop untill you enter any non integer value.
You finished entering all the integer values and then your program will print your list elements.
im working on a problem where i have to obtain all permutations of an arraylist of numbers. The only restriction is that any number cant start with 0, so if we have [0,1,2] we would obtain
[1,2,0]
[1,0,2]
[2,0,1]
[2,1,0]
i know how to do this with 3 loops but the thing is that i have to repeat this to different sets of numbers with differentes sizes, so i need one method that i can apply to different sets of numbers but i have no clue on how to do this. I imagine i have to used some kind of recursive function but i dont know how to implement it so the numbers cant start with a 0. Any ideas? please dont just post the code i want to understand the problem, thank you in advantage!!
Curious question! Interesting code kata.
I naively think I would have a recursive method that takes:
a list of the items currently chosen by the caller
a set of the items available for the callee
The method would iterate over the set to chose 1 more item and call itself with the list extended by this item, and the set reduced by this item. Upon return, remove from list, add back to set and go on with next item (take a defensive copy of the set of course).
If the current list is empty, the selected first item cannot be 0, as per your rules. If you must collect the permutations somewhere (not just print), a 3rd argument would be required for a collection or an observer.
The recursion obvioulsy stops when the available set is empty, at which point the permutation is sent to the collection or observer.
If items can repeat, you may have benefit from sorting them first in order to skip electing the same item again at a given position.
Beware this quires a recursion depth of N, for N items. But the danger is minimal because even with N=10000, it may not stackoverflow, but the CPU time to complete would be order(N!) (probably end of universe...)
You could solve this recursively as described here: Permutation of an ArrayList of numbers using recursion.
The only thing that is missing there is your restriction with the zeros, which could be solved somehow like this (the loop is taken from the example above):
for (List<Integer> al : myLists) {
// The part you need to add:
if (al.get(0) == 0) {
continue;
}
String appender = "";
for (Integer i : al) {
System.out.print(appender + i);
appender = " ";
}
System.out.println();
}
You basically check the first element of each permutation and skip the ones with a leading zero. The continue jumps to the next iteration of the loop and therefore to the next permutation.
The question was asking me to return set containing all the possible combination of strings made up of "cc" and "ddd" for given length n.
so for example if the length given was 5 then set would include "ccddd" and "dddcc".
and length 6 would return set containing "cccccc","dddddd"
and length 7 would return set contating "ccdddcc","dddcccc","ccccddd"
and length 12 will return 12 different combination and so on
However, set returned is empty.
Can you please help?
"Please understand extremeply poor coding style"
public static Set<String> set = new HashSet<String>();
public static Set<String> generateset(int n) {
String s = strings(n,n,"");
return set; // change this
}
public static String strings(int n,int size, String s){
if(n == 3){
s = s + ("cc");
return "";}
if(n == 2){
s = s + ("ddd");
return "";}
if(s.length() == size)
set.add(s);
return strings(n-3,size,s) + strings(n-2,size,s);
}
I think you'll need to rethink your approach. This is not an easy problem, so if you're extremely new to Java (and not extremely familiar with other programming languages), you may want to try some easier problems involving sets, lists, or other collections, before you tackle something like this.
Assuming you want to try it anyway: recursive problems like this require very clear thinking about how you want to accomplish the task. I think you have a general idea, but it needs to be much clearer. Here's how I would approach the problem:
(1) You want a method that returns a list (or set) of strings of length N. Your recursive method returns a single String, and as far as I can tell, you don't have a clear definition of what the resulting string is. (Clear definitions are very important in programming, but probably even more so when solving a complex recursive problem.)
(2) The strings will either begin with "cc" or "ddd". Thus, to form your resulting list, you need to:
(2a) Find all strings of length N-2. This is where you need a recursive call to get all strings of that length. Go through all strings in that list, and add "cc" to the front of each string.
(2b) Similarly, find all strings of length N-3 with a recursive call; go through all the strings in that list, and add "ddd" to the front.
(2c) The resulting list will be all the strings from steps (2a) and (2b).
(3) You need base cases. If N is 0 or 1, the resulting list will be empty. If N==2, it will have just one string, "cc"; if N==3, it will have just one string, "ddd".
You can use a Set instead of a list if you want, since the order won't matter.
Note that it's a bad idea to use a global list or set to hold the results. When a method is calling itself recursively, and every invocation of the method touches the same list or set, you will go insane trying to get everything to work. It's much easier if you let each recursive invocation hold its own local list with the results. Edit: This needs to be clarified. Using a global (i.e. instance field that is shared by all recursive invocations) collection to hold the final results is OK. But the approach I've outlined above involves a lot of intermediate results--i.e. if you want to find all strings whose length is 8, you will also be finding strings whose length is 6, 5, 4, ...; using a global to hold all of those would be painful.
The answer to why set is returned empty is simply follow the logic. Say you execute generateset(5); which will execute strings(5,5,"");:
First iteration strings(5,5,""); : (s.length() == size) is false hence nothing added to set
Second iteration strings(2,5,""); : (n == 2) is true, hence nothing added to set
Third iteration strings(3,5,""); : (n == 3) is true, hence nothing added
to set
So set remains un changed.
I have this code, but it seems pretty unwieldy. Is there a more canonical way of doing so in Java?
public boolean oneDiff(String from, String s) {
if (from.length()!=s.length()) return false;
int differences = 0;
for (int charIndex = 0;charIndex<from.length();charIndex++) {
if (from.charAt(charIndex)!=s.charAt(charIndex)) differences++;
}
return (differences==1);
}
I agree with #mk. However to minimize the loop execution you should not run the loop till the string ends. Instead you can break the loop as soon as the difference becomes greater than 1. Like this:
for (int charIndex = 0;charIndex<from.length();charIndex++) {
if (from.charAt(charIndex)!=s.charAt(charIndex)) differences++;
if(differences > 1) break;
}
return (differences==1);
This will help in faster execution by loop optimization if this is what you want.
Nope, that really is the best way!
There's nothing built-in because this isn't something you need to do often. The closest trick is doing an xor on two integers, and then getting the Hamming Weight using bitCount, in order to check for how many flipped bits they have in common:
Integer.bitCount(int1 ^ int2)
But there's nothing like that for Strings - it's not a common case, so you have to code your own. And the way you've coded it seems fine - you really do have to loop over every character. I guess you could shorten the variable names and remove the parens around your return, but that's just cosmetic.
Am working on some programming homework and am a bit lost. The project is to select the even/odd elements of a listarray and store in another array. It is not the even numbers in each element, but the elements themselves so if an array had values "1,2,5,7,9" and returned the even elements it would give "1, 5, 9". Also have to use recursion. Would anyone be able to give me a starting point or some advice. Though about starting with 2 elements and taking 2nd element and then building up from that, but don't know how it would add on the 2nd pass
public static ArrayList<Integer> even(ArrayList<Integer> list)
ArrayList<Integer> evenlist = ListMethods.deepClone(tList);//make copy of list
if (evenlist.size()<=1) // The list is empty or has one element
{
// return null;// Return the list as is
}
if
(evenlist.size()==2)
{
//return right element
//call method again
//add to list
}
Psuedocode
int[] evens,odds;
function categorize(List<Integer> in,int idx)
if(idx>=in.length)
return
int cur = in[idx]
if(even), add to evens
else add to odds
categorize(in,idx+1)
This sounds similar to the homework I just completed, so if it is (And you're in my class!), I'll not tell you to use any terminology we haven't covered as I know it can be daunting trying to discover something new for practicals (beyond what we have to do).
First, set your exit condition. As you've already said, you have to create a new ArrayList out of the existing one. You are going to remove items from the existing ArrayList, storing the integers that are at even (or odd) indices, until the list is empty.
So your exit condition is:
if (evenList is Empty)
return evenList;
Then, work your way through the steps. I would advise determining if the Array you start with has an even of odd number of steps, something like this:
if (evenList has Even Elements)
int holderForIntsAtEvenElements = last evenList EVEN element
Note we start at the last element, so when you are coming OUT of the recursive method, this will be the last one added to your new ArrayList, and thus it'll be in numerical order. You might find this post interesting to do this: What does this boolean return mean?
We then want to remove the last element from the list and recursively call the method again.
Finally, when we hit our exit condition and start to come out, we want to add the ints we've been storing to them, e.g.:
evenList.add(holderForIntsAtEvenElements);
return evenList;
That doesn't solve one problem, which is what to do with the very first element if the list does NOT have an even number of elements - however, I'll let you try and solve that!
That's a good mix of code and pseudo code and will hopefully help to get you on the right track.
You could use a simple for loop like this:
for (int i = 0; i < list.size(); i += 2) {
System.out.println(list.get(i));
}
If you have to use recursion, here's an outline of the steps you might take. (I won't tell you exactly what to do because you haven't tried anything and it is like homework.)
Take first element and store it
Remove (new) first element from list
Call self