Any shortcut to create a Java array of the first n integers without doing an explicit loop? In R, it would be
intArray = c(1:n)
(and the resulting vector would be 1,2,...,n).
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1Possible duplicate of stackoverflow.com/questions/3790142/….Roberto Reale– Roberto Reale2014-04-17 19:57:31 +00:00Commented Apr 17, 2014 at 19:57
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docs.oracle.com/javase/7/docs/api/java/util/…, int, int)kosa– kosa2014-04-17 19:58:30 +00:00Commented Apr 17, 2014 at 19:58
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You said "array" and "vector", which are 2 different things in Java and the simplest way to do this is different for each of those. Do you know which you want?Jeff Scott Brown– Jeff Scott Brown2014-04-17 19:58:46 +00:00Commented Apr 17, 2014 at 19:58
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1 Answer
If you're using java-8, you could do:
int[] arr = IntStream.range(1, n).toArray();
This will create an array containing the integers from [0, n). You can use rangeClosed if you want to include n in the resulting array.
If you want to specify a step, you could iterate and then limit the stream to take the first n elements you want.
int[] arr = IntStream.iterate(0, i ->i + 2).limit(10).toArray(); //[0, 2, 4, 6, 8, 10, 12, 14, 16, 18]
Otherwise I guess the simplest way to do is to use a loop and fill the array. You can create a helper method if you want.
static int[] fillArray(int from, int to, int step){
if(to < from || step <= 0)
throw new IllegalArgumentException("to < from or step <= 0");
int[] array = new int[(to-from)/step+1];
for(int i = 0; i < array.length; i++){
array[i] = from;
from += step;
}
return array;
}
...
int[] arr3 = fillArray(0, 10, 3); //[0, 3, 6, 9]
You can adapt this method as your needs to go per example from an upperbound to a lowerbound with a negative step.
