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

Question

I have a String[] with values like so:

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

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

Answer 1

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);

Answer 2

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.)

Answer 3

You can use ArrayUtils.contains from Apache Commons Lang

public static boolean contains(Object[] array, Object objectToFind)

Note that this method returns false if the passed array is null.

There are also methods available for primitive arrays of all kinds.

Example:

String[] fieldsToInclude = { "id", "name", "location" };

if ( ArrayUtils.contains( fieldsToInclude, "id" ) ) {
    // Do some stuff.
}

Answer 4

Just simply implement it by hand:

public static <T> boolean contains(final T[] array, final T v) {
    for (final T e : array)
        if (e == v || v != null && v.equals(e))
            return true;

    return false;
}

Improvement:

The v != null condition is constant inside the method. It always evaluates to the same Boolean value during the method call. So if the input array is big, it is more efficient to evaluate this condition only once, and we can use a simplified/faster condition inside the for loop based on the result. The improved contains() method:

public static <T> boolean contains2(final T[] array, final T v) {
    if (v == null) {
        for (final T e : array)
            if (e == null)
                return true;
    } 
    else {
        for (final T e : array)
            if (e == v || v.equals(e))
                return true;
    }

    return false;
}

Answer 5

Four Different Ways to Check If an Array Contains a Value

1. Using List:

   public static boolean useList(String[] arr, String targetValue) {
       return Arrays.asList(arr).contains(targetValue);
   }
   

2. Using Set:

   public static boolean useSet(String[] arr, String targetValue) {
       Set<String> set = new HashSet<String>(Arrays.asList(arr));
       return set.contains(targetValue);
   }
   

3. Using a simple loop:

   public static boolean useLoop(String[] arr, String targetValue) {
       for (String s: arr) {
           if (s.equals(targetValue))
               return true;
       }
       return false;
   }
   

4. Using Arrays.binarySearch():

   The code below is wrong, it is listed here for completeness. binarySearch() can ONLY be used on sorted arrays. You will find the result is weird below. This is the best option when array is sorted.

   public static boolean binarySearch(String[] arr, String targetValue) {  
       return Arrays.binarySearch(arr, targetValue) >= 0;
   }
   

Quick Example:

String testValue="test";
String newValueNotInList="newValue";
String[] valueArray = { "this", "is", "java" , "test" };
Arrays.asList(valueArray).contains(testValue); // returns true
Arrays.asList(valueArray).contains(newValueNotInList); // returns false

Answer 6

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.

Answer 7

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));
    }
}

Answer 8

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");

Answer 9

With Java 8 you can create a stream and check if any entries in the stream matches "s":

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

Or as a generic method:

public static <T> boolean arrayContains(T[] array, T value) {
    return Arrays.stream(array).anyMatch(value::equals);
}

Answer 10

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.

Answer 11

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;
}

Answer 12

Actually, if you use HashSet<String> as Tom Hawtin proposed you don't need to worry about sorting, and your speed is the same as with binary search on a presorted array, probably even faster.

It all depends on how your code is set up, obviously, but from where I stand, the order would be:

On an unsorted array:

1. HashSet
2. asList
3. sort & binary

On a sorted array:

1. HashSet
2. Binary
3. asList

So either way, HashSet for the win.

Answer 13

If you have the google collections library, Tom's answer can be simplified a lot by using ImmutableSet (http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/ImmutableSet.html)

This really removes a lot of clutter from the initialization proposed

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

Answer 14

One possible solution:

import java.util.Arrays;
import java.util.List;

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

  public static void main(String args[]) {

      if (VALUES.contains("AB")) {
          System.out.println("Contains");
      } else {
          System.out.println("Not contains");
      }
  }
}

Answer 15

Developers often do:

Set<String> set = new HashSet<String>(Arrays.asList(arr));
return set.contains(targetValue);

The above code works, but there is no need to convert a list to set first. Converting a list to a set requires extra time. It can as simple as:

Arrays.asList(arr).contains(targetValue);

or

for (String s : arr) {
    if (s.equals(targetValue))
        return true;
}

return false;

The first one is more readable than the second one.

Answer 16

In Java 8 use Streams.

List<String> myList =
        Arrays.asList("a1", "a2", "b1", "c2", "c1");

myList.stream()
        .filter(s -> s.startsWith("c"))
        .map(String::toUpperCase)
        .sorted()
        .forEach(System.out::println);

Answer 17

Using a simple loop is the most efficient way of doing this.

boolean useLoop(String[] arr, String targetValue) {
    for(String s: arr){
        if(s.equals(targetValue))
            return true;
    }
    return false;
}

Courtesy to Programcreek

Answer 18

1. For arrays of limited length use the following (as given by camickr). This is slow for repeated checks, especially for longer arrays (linear search).

    Arrays.asList(...).contains(...)
   

2. For fast performance if you repeatedly check against a larger set of elements

   • An array is the wrong structure. Use a TreeSet and add each element to it. It sorts elements and has a fast exist() method (binary search).

   • If the elements implement Comparable & you want the TreeSet sorted accordingly:

     ElementClass.compareTo() method must be compatable with ElementClass.equals(): see Triads not showing up to fight? (Java Set missing an item)

     TreeSet myElements = new TreeSet();
     
     // Do this for each element (implementing *Comparable*)
     myElements.add(nextElement);
     
     // *Alternatively*, if an array is forceably provided from other code:
     myElements.addAll(Arrays.asList(myArray));
     

   • Otherwise, use your own Comparator:

     class MyComparator implements Comparator<ElementClass> {
          int compareTo(ElementClass element1; ElementClass element2) {
               // Your comparison of elements
               // Should be consistent with object equality
          }
     
          boolean equals(Object otherComparator) {
               // Your equality of comparators
          }
     }
     
     
     // construct TreeSet with the comparator
     TreeSet myElements = new TreeSet(new MyComparator());
     
     // Do this for each element (implementing *Comparable*)
     myElements.add(nextElement);
     

   • The payoff: check existence of some element:

     // Fast binary search through sorted elements (performance ~ log(size)):
     boolean containsElement = myElements.exists(someElement);
     

Answer 19

Use the following (the contains() method is ArrayUtils.in() in this code):

ObjectUtils.java

public class ObjectUtils {
    /**
     * A null safe method to detect if two objects are equal.
     * @param object1
     * @param object2
     * @return true if either both objects are null, or equal, else returns false.
     */
    public static boolean equals(Object object1, Object object2) {
        return object1 == null ? object2 == null : object1.equals(object2);
    }
}

ArrayUtils.java

public class ArrayUtils {
    /**
     * Find the index of of an object is in given array,
     * starting from given inclusive index.
     * @param ts    Array to be searched in.
     * @param t     Object to be searched.
     * @param start The index from where the search must start.
     * @return Index of the given object in the array if it is there, else -1.
     */
    public static <T> int indexOf(final T[] ts, final T t, int start) {
        for (int i = start; i < ts.length; ++i)
            if (ObjectUtils.equals(ts[i], t))
                return i;
        return -1;
    }

    /**
     * Find the index of of an object is in given array, starting from 0;
     * @param ts Array to be searched in.
     * @param t  Object to be searched.
     * @return indexOf(ts, t, 0)
     */
    public static <T> int indexOf(final T[] ts, final T t) {
        return indexOf(ts, t, 0);
    }

    /**
     * Detect if the given object is in the given array.
     * @param ts Array to be searched in.
     * @param t  Object to be searched.
     * @return If indexOf(ts, t) is greater than -1.
     */
    public static <T> boolean in(final T[] ts, final T t) {
        return indexOf(ts, t) > -1;
    }
}

As you can see in the code above, that there are other utility methods ObjectUtils.equals() and ArrayUtils.indexOf(), that were used at other places as well.

Answer 20

Try this:

ArrayList<Integer> arrlist = new ArrayList<Integer>(8);

// use add() method to add elements in the list
arrlist.add(20);
arrlist.add(25);
arrlist.add(10);
arrlist.add(15);

boolean retval = arrlist.contains(10);
if (retval == true) {
    System.out.println("10 is contained in the list");
}
else {
    System.out.println("10 is not contained in the list");
}

Answer 21

Arrays.asList() -> then calling the contains() method will always work, but a search algorithm is much better since you don't need to create a lightweight list wrapper around the array, which is what Arrays.asList() does.

public boolean findString(String[] strings, String desired){
   for (String str : strings){
       if (desired.equals(str)) {
           return true;
       }
   }
   return false; //if we get here… there is no desired String, return false.
}

Answer 22

If you don't want it to be case sensitive

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

Answer 23

Check this

String[] VALUES = new String[]{"AB", "BC", "CD", "AE"};
String s;

for (int i = 0; i < VALUES.length; i++) {
    if (VALUES[i].equals(s)) {
        // do your stuff
    } else {
        //do your stuff
    }
}

Answer 24

Use Array.BinarySearch(array,obj) for finding the given object in array or not.

Example:

if (Array.BinarySearch(str, i) > -1)` → true --exists

false --not exists

Answer 25

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

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

Answer 26

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.

Answer 27

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

Answer 28

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

Answer 29

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));
}