How do I determine whether an array contains a particular value in Java?

Arrays.asList(yourArray).contains(yourValue)

Warning: this doesn't work for arrays of primitives (see the comments).

Since java-8 you can now use Streams.

String[] values = {"AB","BC","CD","AE"};
boolean contains = Arrays.stream(values).anyMatch("s"::equals);

To check whether an array of int, double or long contains a value use IntStream, DoubleStream or LongStream respectively.

Example

int[] a = {1,2,3,4};
boolean contains = IntStream.of(a).anyMatch(x -> x == 4);

Concise update for Java SE 9

Reference arrays are bad. For this case we are after a set. Since Java SE 9 we have Set.of.

private static final Set<String> VALUES = Set.of(
    "AB","BC","CD","AE"
);

"Given String s, is there a good way of testing whether VALUES contains s?"

VALUES.contains(s)

O(1).

The right type, immutable, O(1) and concise. Beautiful.*

Original answer details

Just to clear the code up to start with. We have (corrected):

public static final String[] VALUES = new String[] {"AB","BC","CD","AE"};

This is a mutable static which FindBugs will tell you is very naughty. Do not modify statics and do not allow other code to do so also. At an absolute minimum, the field should be private:

private static final String[] VALUES = new String[] {"AB","BC","CD","AE"};

(Note, you can actually drop the new String[]; bit.)

Reference arrays are still bad and we want a set:

private static final Set<String> VALUES = new HashSet<String>(Arrays.asList(
     new String[] {"AB","BC","CD","AE"}
));

(Paranoid people, such as myself, may feel more at ease if this was wrapped in Collections.unmodifiableSet - it could then even be made public.)

(*To be a little more on brand, the collections API is predictably still missing immutable collection types and the syntax is still far too verbose, for my tastes.)

If the array is not sorted, you will have to iterate over everything and make a call to equals on each.

If the array is sorted, you can do a binary search, there's one in the Arrays class.

Generally speaking, if you are going to do a lot of membership checks, you may want to store everything in a Set, not in an array.

For what it's worth I ran a test comparing the 3 suggestions for speed. I generated random integers, converted them to a String and added them to an array. I then searched for the highest possible number/string, which would be a worst case scenario for the asList().contains().

When using a 10K array size the results were:

Sort & Search   : 15
Binary Search   : 0
asList.contains : 0

When using a 100K array the results were:

Sort & Search   : 156
Binary Search   : 0
asList.contains : 32

So if the array is created in sorted order the binary search is the fastest, otherwise the asList().contains would be the way to go. If you have many searches, then it may be worthwhile to sort the array so you can use the binary search. It all depends on your application.

I would think those are the results most people would expect. Here is the test code:

import java.util.*;

public class Test {
    public static void main(String args[]) {
        long start = 0;
        int size = 100000;
        String[] strings = new String[size];
        Random random = new Random();

        for (int i = 0; i < size; i++)
            strings[i] = "" + random.nextInt(size);

        start = System.currentTimeMillis();
        Arrays.sort(strings);
        System.out.println(Arrays.binarySearch(strings, "" + (size - 1)));
        System.out.println("Sort & Search : "
                + (System.currentTimeMillis() - start));

        start = System.currentTimeMillis();
        System.out.println(Arrays.binarySearch(strings, "" + (size - 1)));
        System.out.println("Search        : "
                + (System.currentTimeMillis() - start));

        start = System.currentTimeMillis();
        System.out.println(Arrays.asList(strings).contains("" + (size - 1)));
        System.out.println("Contains      : "
                + (System.currentTimeMillis() - start));
    }
}

Instead of using the quick array initialisation syntax too, you could just initialise it as a List straight away in a similar manner using the Arrays.asList method, e.g.:

public static final List<String> STRINGS = Arrays.asList("firstString", "secondString" ...., "lastString");

Then you can do (like above):

STRINGS.contains("the string you want to find");

You can use the Arrays class to perform a binary search for the value. If your array is not sorted, you will have to use the sort functions in the same class to sort the array, then search through it.

ObStupidAnswer (but I think there's a lesson in here somewhere):

enum Values {
    AB, BC, CD, AE
}

try {
    Values.valueOf(s);
    return true;
} catch (IllegalArgumentException exc) {
    return false;
}

If you don't want it to be case sensitive

Arrays.stream(VALUES).anyMatch(s::equalsIgnoreCase);

the shortest solution
the array VALUES may contain duplicates
since Java 9

List.of(VALUES).contains(s);

Try using Java 8 predicate test method

Here is a full example of it.

import java.util.Arrays;
import java.util.List;
import java.util.function.Predicate;

public class Test {
    public static final List<String> VALUES =
            Arrays.asList("AA", "AB", "BC", "CD", "AE");

    public static void main(String args[]) {
        Predicate<String> containsLetterA = VALUES -> VALUES.contains("AB");
        for (String i : VALUES) {
            System.out.println(containsLetterA.test(i));
        }
    }
}

http://mytechnologythought.blogspot.com/2019/10/java-8-predicate-test-method-example.html

https://github.com/VipulGulhane1/java8/blob/master/Test.java

Create a boolean initially set to false. Run a loop to check every value in the array and compare to the value you are checking against. If you ever get a match, set boolean to true and stop the looping. Then assert that the boolean is true.

As I'm dealing with low level Java using primitive types byte and byte[], the best so far I got is from bytes-java https://github.com/patrickfav/bytes-java seems a fine piece of work

You can check it by two methods

A) By converting the array into string and then check the required string by .contains method

String a = Arrays.toString(VALUES);
System.out.println(a.contains("AB"));
System.out.println(a.contains("BC"));
System.out.println(a.contains("CD"));
System.out.println(a.contains("AE"));

B) This is a more efficent method

Scanner s = new Scanner(System.in);

String u = s.next();
boolean d = true;
for (int i = 0; i < VAL.length; i++) {
    if (VAL[i].equals(u) == d)
        System.out.println(VAL[i] + " " + u + VAL[i].equals(u));
}