Find an integer in an Array in a short manner in Java

Question

I already tried this post: Java, Simplified check if int array contains int

and tried this

int[] temp = {3,9,15,21,27,33,39};
HashSet<Integer> otherBy3 = new HashSet<Integer>(Arrays.asList(temp));

based on the post i saw on the above link instructing me to do this:

HashSet<Integer> set= new HashSet<Integer>(Arrays.asList(intArray));
set.contains(intValue)

but I keep getting this error

cannot find symbol
symbol  : constructor HashSet(java.util.List<int[]>)
location: class java.util.HashSet<java.lang.Integer> HashSet<Integer> otherBy3 = new HashSet<Integer>(Arrays.asList(temp));

Any help would be appreciated

The first answer on that link points out what happens when `int[]` to `Arrays.asList` (and why it doesn't do what you want)... — Dennis Meng, Jan 17 '14 at 07:51

score 2 · Accepted Answer · answered Jan 17 '14 at 07:50

2

Change

int[] temp = ...

to

Integer[] temp = ...

answered Jan 17 '14 at 07:50

Moritz Petersen

12,902
3
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score 2 · Answer 2 · answered Jan 17 '14 at 07:52

2

Integer[] temp = {3,9,15,21,27,33,39};
HashSet<Integer> set= new HashSet<Integer>(Arrays.asList(temp));
System.out.println(set.contains(3));

answered Jan 17 '14 at 07:52

Vinayak Pingale

1,315
8
19

score 1 · Answer 3 · answered Jan 17 '14 at 07:52

1

Change int to Integer because you are declaring the List with generic as Integer

Integer[] temp = new Integer[] { 3, 9, 15, 21, 27, 33, 39 };
HashSet<Integer> otherBy3 = new HashSet<Integer>(Arrays.asList(temp));
otherBy3.contains(13);

answered Jan 17 '14 at 07:52

Jagadesh Seeram

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wmorrison365 · Answer 4 · 2014-01-17T11:10:20.047

Asusming the real problem ou're asking about is "how to find an integer in an int[] array", why not just use Arrays.html#binarySearch. It's a one-liner, very quick O( log n) and is the common way of doing it. The only pre-req is that your list is sorted and yours appears to be (isn't it?).

int[] temp = {3,9,15,21,27,33,39};
int k = 28;    //Not present in temp
if (Arrays.binarySearch(temp, k) > 0) {
    //key found
}
else {
    //key not found - returns -5 (negative indicating not found)
}

score 0 · Answer 5 · answered Jan 17 '14 at 10:18

0

I find it somewhat disturbing that constructing several temporary objects is regarded "shorter" than a simple freaking loop:

 int[] temp = {3,9,15,21,27,33,39};
 boolean found = false;
 for (int i=0; i<temp.length && !found; ++i) {
     found = valueToFind == temp[i];         
 }

Its about the same amount of code as the high overhead approaches.

answered Jan 17 '14 at 10:18

Durandal

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4
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What about a [binary search](http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=binarySearch) instead? Takes at most log n. You even get a free impl in `Arrays#binarySearch`. – wmorrison365 Jan 17 '14 at 11:13
That requires sorting. Sorting is O(N log N) (at best) vs a plain O(N) for the simple loop. The question is mainly about "short" (in amount of code lines, I assume), and I just put this answer to point out that "just doing it" is about as long as going through lengths to do it as a byproduct of something else. If the aim would be on performance, you'd need a lot more context to select the best method. – Durandal Jan 17 '14 at 11:18
OK. The list is sorted in the question but perhaps that's just incidental. – wmorrison365 Jan 17 '14 at 11:37

Find an integer in an Array in a short manner in Java

5 Answers5