Java, Simplified check if int array contains int

Question

Basically my mate has been saying that I could make my code shorter by using a different way of checking if an int array contains an int, although he won't tell me what it is :P.

Current:

public boolean contains(final int[] array, final int key) {
    for (final int i : array) {
        if (i == key) {
            return true;
        }
    }
    return false;
}

Have also tried this, although it always returns false for some reason.

public boolean contains(final int[] array, final int key) {
    return Arrays.asList(array).contains(key);
}

Could anyone help me out?

Thank you.

Your Arrays.asList(...) call takes a vararg, that is it will wrap the arbitrary number of arguments you might pass into that in a List. In your case, you're getting a list of arrays with a single element, and this list obviously does not contain the int. — sarcan
– sarcan, Commented Aug 18, 2012 at 16:45
check Hashset based retrial mechanism answer. It is the fastest way. — Amit Deshpande
– Amit Deshpande, Commented Aug 18, 2012 at 17:22
I don't see any point at making your original code shorter since your argument is a primitive array and your code is very clear and straightfoward. ArrayList implementation is doing the same. — Genzer
– Genzer, Commented Aug 18, 2012 at 17:38
I would not make your code shorter. (1) arraylist does the same thing you did. (2) - more important stuff is that the shorten code using Arrays.asList creates new object, which could be problem within some performance critical code. The first code snippet is best thing you can do. — Martin Podval
– Martin Podval, Commented Jan 5, 2015 at 6:33

Donald Duck is with Ukraine · Accepted Answer · 2021-09-02 18:44:38Z

84

You could simply use ArrayUtils.contains from Apache Commons Lang library.

public boolean contains(final int[] array, final int key) {     
    return ArrayUtils.contains(array, key);
}

edited Sep 2, 2021 at 18:44

Donald Duck is with Ukraine

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answered Aug 18, 2012 at 16:50

Reimeus

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4 Comments

mejdev Over a year ago

As long as you are using ArrayUtils, is there any reason not to use ArrayUtils.contains

Reimeus Over a year ago

No reason whatsoever :)

Krzysiek Over a year ago

It's worth noting that ArrayUtils.contains() is part of Apache Commons Lang library. Even though that's a great lib, it is probably still not a good idea to add external dependency just to check if array contains an element :D

TriCore Over a year ago

ArrayUtils is a thing of past. Java 8+ and Guava have pretty amazing goodies!!

TriCore · Accepted Answer · 2015-12-31 05:52:06Z

71

Here is Java 8 solution

public static boolean contains(final int[] arr, final int key) {
    return Arrays.stream(arr).anyMatch(i -> i == key);
}

answered Dec 31, 2015 at 5:52

TriCore

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2 Comments

Mathieu Over a year ago

Note that this call requires API level 24

TXN Sep 27 at 17:02

I feel uneasy about creating several temporary objects (Spliterator, Stream) in a method hierarchy several levels deep just to iterate over an array.

dan1st · Accepted Answer · 2021-03-09 12:30:09Z

40

It's because Arrays.asList(array) returns List<int[]>. The array argument is treated as one value you want to wrap (you get a list of arrays of ints), not as vararg.

Note that it does work with object types (not primitives):

public boolean contains(final String[] array, final String key) {
    return Arrays.asList(array).contains(key);
}

or even:

public <T>  boolean contains(final T[] array, final T key) {
    return Arrays.asList(array).contains(key);
}

But you cannot have List<int> and autoboxing is not working here.

edited Mar 9, 2021 at 12:30

dan1st

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answered Aug 18, 2012 at 16:45

Tomasz Nurkiewicz

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1 Comment

subhashis Over a year ago

Why autoboxing does not work, is it because it is declared as final?

Evert · Accepted Answer · 2013-07-16 16:39:15Z

21

Guava offers additional methods for primitive types. Among them a contains method which takes the same arguments as yours.

public boolean contains(final int[] array, final int key) {
    return Ints.contains(array, key);
}

You might as well statically import the guava version.

See Guava Primitives Explained

answered Jul 16, 2013 at 16:39

Evert

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Comments

Martin Marconcini · Accepted Answer · 2017-02-27 21:57:54Z

21

A different way:

public boolean contains(final int[] array, final int key) {  
     Arrays.sort(array);  
     return Arrays.binarySearch(array, key) >= 0;  
}

This modifies the passed-in array. You would have the option to copy the array and work on the original array i.e. int[] sorted = array.clone();
But this is just an example of short code. The runtime is O(NlogN) while your way is O(N)

edited Feb 27, 2017 at 21:57

Martin Marconcini

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answered Aug 18, 2012 at 16:50

Cratylus

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6 Comments

Zong Over a year ago

I think I'd be surprised if a contains method modified my array.

Cratylus Over a year ago

@ZongLi:This is just an example for the OP.Updated OP if we are nitpicking

icza Over a year ago

From javadoc of binarySearch(): "the return value will be >= 0 if and only if the key is found." so Arrays.binarySearch(array,key)>=0 should be returned!

icza Over a year ago

Supplement: The return value of binarySearch() is (-(insertion point) - 1) if key is not contained which may likely be a value other than -1.

TriCore Over a year ago

Sorting and binary-searching is more "expensive" than full scan. (Besides the fact that array would be modified)

|

Willy Wonka · Accepted Answer · 2014-12-16 13:13:39Z

19

I know it's super late, but try Integer[] instead of int[].

answered Dec 16, 2014 at 13:13

Willy Wonka

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Comments

JaskeyLam · Accepted Answer · 2014-08-22 08:54:51Z

3

1.one-off uses

List<T> list=Arrays.asList(...)
list.contains(...)

2.use HashSet for performance consideration if you use more than once.

Set <T>set =new HashSet<T>(Arrays.asList(...));
set.contains(...)

answered Aug 22, 2014 at 8:54

JaskeyLam

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Comments

Hetal Rachh · Accepted Answer · 2019-05-26 13:15:34Z

2

You can convert your primitive int array into an arraylist of Integers using below Java 8 code,

List<Integer> arrayElementsList = Arrays.stream(yourArray).boxed().collect(Collectors.toList());

And then use contains() method to check if the list contains a particular element,

boolean containsElement = arrayElementsList.contains(key);

answered May 26, 2019 at 13:15

Hetal Rachh

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Comments

Mahmoud · Accepted Answer · 2021-11-19 01:48:25Z

2

Java 9+

public boolean contains(final int[] array, final int key) {
    return List.of(array).contains(key);
}

answered Nov 19, 2021 at 1:48

Mahmoud

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1 Comment

TXN Sep 27 at 17:06

This compiles, but it doesn't work. It creates a List<int[]> with one element that will never equal the key, so your method always returns false.

IgniteCoders · Accepted Answer · 2018-04-05 14:49:58Z

1

You can use java.util.Arrays class to transform the array T[?] in a List<T> object with methods like contains:

Arrays.asList(int[] array).contains(int key);

edited Apr 5, 2018 at 14:49

IgniteCoders

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answered Nov 20, 2014 at 4:14

Mourad El Aomari

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1 Comment

TXN Sep 27 at 17:08

This compiles, but it doesn't work. It creates a List<int[]>, not a List<Integer> and will thus always return false.

Farhad Baghirov · Accepted Answer · 2017-05-15 05:42:26Z

0

this worked in java 8

public static boolean contains(final int[] array, final int key)
{
return Arrays.stream(array).anyMatch(n->n==key);
}

edited May 15, 2017 at 5:42

answered Oct 29, 2016 at 19:44

Farhad Baghirov

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5 Comments

TriCore Over a year ago

It should immediately return on first match, instead this will still scan all the items in array, even if it did find the match. (Consider an array of trilion items)

Farhad Baghirov Over a year ago

You are right try this public static boolean contains(final int[] array, final int key) { return Arrays.stream(array).anyMatch(n->n==key); }

LordParsley Over a year ago

Java 8 stream anyMatch is a short-circuit operation and will not scan all items in the array.

Farhad Baghirov Over a year ago

@LordParsley Above code's goal is check element in array ,not scan all element of array.

LordParsley Over a year ago

Sorry, I see the answer was edited. I was just reiterating it was correct as it will not need to scan all if it finds one part way.

Mosius · Accepted Answer · 2018-12-18 14:36:27Z

0

Try this:

public static void arrayContains(){
    int myArray[]={2,2,5,4,8};

    int length=myArray.length;

    int toFind = 5;
    boolean found = false;

    for(int i = 0; i < length; i++) {
        if(myArray[i]==toFind) {
            found=true;
        }
    }

    System.out.println(myArray.length);
    System.out.println(found); 
}

edited Dec 18, 2018 at 14:36

Mosius

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answered Dec 18, 2018 at 13:33

Elavarasan S

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Comments

Kunal · Accepted Answer · 2021-06-16 14:48:08Z

0

private static void solutions() {
    int[] A = { 1, 5, 10, 20, 40, 80 };
    int[] B = { 6, 7, 20, 80, 100 };
    int[] C = { 3, 4, 15, 20, 30, 70, 80, 120 };

    List<Integer> aList = Arrays.stream(A).boxed().collect(Collectors.toList());

    List<Integer> cList = Arrays.stream(C).boxed().collect(Collectors.toList());
    String s = "";
    for (Integer a : C) {
        if (aList.contains(a) && cList.contains(a)) {
            s = s.concat(String.valueOf(a)).concat("->");
        }
    }
}

answered Jun 16, 2021 at 14:48

Kunal

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Comments

Kaveh Ghahremani · Accepted Answer · 2012-08-18 17:18:22Z

Depending on how large your array of int will be, you will get much better performance if you use collections and .contains rather than iterating over the array one element at a time:

import static org.junit.Assert.assertTrue;
import java.util.HashSet;

import org.junit.Before;
import org.junit.Test;

public class IntLookupTest {

int numberOfInts = 500000;
int toFind = 200000;
int[] array;

HashSet<Integer> intSet;

@Before
public void initializeArrayAndSet() {
    array = new int[numberOfInts];
    intSet = new HashSet<Integer>();
    for(int i = 0; i < numberOfInts; i++) {
        array[i] = i;
        intSet.add(i);
    }
}

@Test
public void lookupUsingCollections() {
    assertTrue(intSet.contains(toFind));
}

@Test
public void iterateArray() {
    assertTrue(contains(array, toFind));

}

public boolean contains(final int[] array, final int key) {
    for (final int i : array) {
        if (i == key) {
            return true;
        }
    }
    return false;
}
}

Kumar Vivek Mitra · Accepted Answer · 2012-08-18 17:15:42Z

-6

Try Integer.parseInt() to do this.....

public boolean chkInt(final int[] array){
    int key = false;

    for (Integer i : array){


          try{

                   Integer.parseInt(i);
                   key = true;
                   return key;

             }catch(NumberFormatException ex){

                   key = false;

                   return key;

              }


     }
}

answered Aug 18, 2012 at 17:15

Kumar Vivek Mitra

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