I have this code:
public static String SelectRandomFromTemplate(String template,int count) { String[] split = template.split("|"); List<String> list=Arrays.asList(split); Random r = new Random(); while( list.size() > count ) { list.remove(r.nextInt(list.size())); } return StringUtils.join(list, ", "); }
I get this:
06-03 15:05:29.614: ERROR/AndroidRuntime(7737): java.lang.UnsupportedOperationException 06-03 15:05:29.614: ERROR/AndroidRuntime(7737): at java.util.AbstractList.remove(AbstractList.java:645)
How would be this the correct way? Java.15
public class UnsupportedOperationException extends RuntimeException. Thrown to indicate that the requested operation is not supported. This class is a member of the Java Collections Framework.
remove(int index) method removes the element at the specified position in this list. Shifts any subsequent elements to the left (subtracts one from their indices).
Quite a few problems with your code:
Arrays.asList
returning a fixed-size listFrom the API:
Arrays.asList
: Returns a fixed-size list backed by the specified array.
You can't add
to it; you can't remove
from it. You can't structurally modify the List
.
Create a LinkedList
, which supports faster remove
.
List<String> list = new LinkedList<String>(Arrays.asList(split));
split
taking regexFrom the API:
String.split(String regex)
: Splits this string around matches of the given regular expression.
|
is a regex metacharacter; if you want to split on a literal |
, you must escape it to \|
, which as a Java string literal is "\\|"
.
template.split("\\|")
Instead of calling remove
one at a time with random indices, it's better to generate enough random numbers in the range, and then traversing the List
once with a listIterator()
, calling remove()
at appropriate indices. There are questions on stackoverflow on how to generate random but distinct numbers in a given range.
With this, your algorithm would be O(N)
.
This one has burned me many times. Arrays.asList
creates an unmodifiable list. From the Javadoc: Returns a fixed-size list backed by the specified array.
Create a new list with the same content:
newList.addAll(Arrays.asList(newArray));
This will create a little extra garbage, but you will be able to mutate it.
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