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Checking if number is a member of a matrix [duplicate]

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?

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.)

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.
}

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;
}

Four Different Ways to Check If an Array Contains a Value
Using List:
public static boolean useList(String[] arr, String targetValue) {
return Arrays.asList(arr).contains(targetValue);
}
Using Set:
public static boolean useSet(String[] arr, String targetValue) {
Set<String> set = new HashSet<String>(Arrays.asList(arr));
return set.contains(targetValue);
}
Using a simple loop:
public static boolean useLoop(String[] arr, String targetValue) {
for (String s: arr) {
if (s.equals(targetValue))
return true;
}
return false;
}
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

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");

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);
}

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;
}

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:
HashSet
asList
sort & binary
On a sorted array:
HashSet
Binary
asList
So either way, HashSet for the win.

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.

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");

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);

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");
}
}
}

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

the shortest solution
the array VALUES may contain duplicates
since Java 9
List.of(VALUES).contains(s);

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.

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(...)
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);

If you don't want it to be case sensitive
Arrays.stream(VALUES).anyMatch(s::equalsIgnoreCase);

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");
}

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
}
}

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.
}

Use below -
String[] values = {"AB","BC","CD","AE"};
String s = "A";
boolean contains = Arrays.stream(values).anyMatch(v -> v.contains(s));

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

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));
}

Sorting java arrayList with predefined order

I have an unsorted list but I want to sort in a custom way i.e.
item_one_primary.pls
item_one_secondary.pls
item_one_last.pls
item_two_last.pls
item_two_primary.pls
item_two_secondary.pls
item_three_secondary.pls
item_three_last.pls
item_three_primary.pls
Here is my predefined order : primary, secondary, last
Above unordered list once the ordering is applied should look like this :
item_one_primary.pls
item_one_secondary.pls
item_one_last.pls
item_two_primary.pls
item_two_secondary.pls
item_two_last.pls
item_three_primary.pls
item_three_secondary.pls
item_three_last.pls
I tried something with comparator but I end up something like this :
item_one_primary.pls
item_two_primary.pls
item_three_primary.pls
...
Does anyone have an idea how to get this sorted?
Here is some code I've used :
List<String> predefinedOrder;
public MyComparator(String[] predefinedOrder) {
this.predefinedOrder = Arrays.asList(predefinedOrder);
}
#Override
public int compare(String item1, String item2) {
return predefinedOrder.indexOf(item1) - predefinedOrder.indexOf(item2);
}
I didn't include the splits(first split by dot(.) second split by underscore(_) to get the item in pre-ordered list).

You have to use a Comparator that checks first the item number and only if they are equal, check your predefined order.
Try something like this:
public int compare(Object o1, Object o2) {
String s1 = (String) o1;
String s2 = (String) o2;
String[] a1 = s1.split("_");
String[] a2 = s2.split("_");
/* If the primary elements of order are equal the result is
the order of the second elements of order */
if (a1[1].compareTo(a2[1]) == 0) {
return a1[2].compareTo(a2[2]);
/* If they are not equal, we just order by the primary elements */
} else {
return a1[1].compareTo(a2[1]);
}
}
This is just a basic example, some extra error checking would be nice.

A solution using the Google Guava API yields a simple and readable result:
// some values
List<String> list = Lists.newArrayList("item_one_primary", "item_one_secondary", "item_one_last");
// define an explicit ordering that uses the result of a function over the supplied list
Ordering o = Ordering.explicit("primary", "secondary", "last").onResultOf(new Function<String, String>() {
// the function splits a values by '_' and uses the last element (primary, secondary etc.)
public String apply(String input) {
return Lists.newLinkedList(Splitter.on("_").split(input)).getLast();
}
});
// the ordered result
System.out.println("o.sortedCopy(list); = " + o.sortedCopy(list));

List Sorting puzzle

Assuming I have
final Iterable<String> unsorted = asList("FOO", "BAR", "PREFA", "ZOO", "PREFZ", "PREFOO");
What can I do to transform this unsorted list into this:
[PREFZ, PREFA, BAR, FOO, PREFOO, ZOO]
(a list which begin with known values that must appears first (here "PREFA" and "PREFZ") and the rest is alphabetically sorted)
I think there are some usefull classes in guava that can make the job (Ordering, Predicates...), but I have not yet found a solution...

I would keep separate lists.
One for known values and unknown values. And sort them separately, when you need them in a one list you can just concatenate them.
knownUnsorted.addAll(unsorted.size - 1, unknonwUnsorted);

I suggest filling List with your values and using Collections.sort(...).
Something like
Collections.sort(myList, new FunkyComparator());
using this:
class FunkyComparator implements Comparator {
private static Map<String,Integer> orderedExceptions =
new HashMap<String,Integer>(){{
put("PREFZ", Integer.valueOf(1));
put("PREFA", Integer.valueOf(2));
}};
public int compare(Object o1, Object o2) {
String s1 = (String) o1;
String s2 = (String) o2;
Integer i1 = orderedExceptions.get(s1);
Integer i2 = orderedExceptions.get(s2);
if (i1 != null && i2 != null) {
return i1 - i2;
}
if (i1 != null) {
return -1;
}
if (i2 != null) {
return +1;
}
return s1.compareTo(s2);
}
}

Note: This is not the most efficient solution. It is just a simple, straightforward solution that gets the job done.
I would first use Collections.sort(list) to sort the list.
Then, I would remove the known items, and add them to the front.
String special = "PREFA";
if (list.remove(special)
list.add(0, special);
Or, if you have a list of array of these values you need in the front you could do:
String[] knownValues = {};
for (String s: knownValues) {
if (list.remove(s))
list.add(0, s);
}

Since I'm a fan of the guava lib, I wanted to find a solution using it. I don't know if it's efficient, neither if you find it as simple as others solution, but it's here:
final Iterable<String> all = asList("FOO", "BAR", "PREFA", "ZOO", "PREFOO", "PREFZ");
final List<String> mustAppearFirst = asList("PREFZ", "PREFA");
final Iterable<String> sorted =
concat(
Ordering.explicit(mustAppearFirst).sortedCopy(filter(all, in(mustAppearFirst))),
Ordering.<String>natural().sortedCopy(filter(all, not(in(mustAppearFirst)))));

You specifically mentioned guava; along with Sylvain M's answer, here's another way (more as an academic exercise and demonstration of guava's flexibility than anything else)
// List is not efficient here; for large problems, something like SkipList
// is more suitable
private static final List<String> KNOWN_INDEXES = asList("PREFZ", "PREFA");
private static final Function<Object, Integer> POSITION_IN_KNOWN_INDEXES
= new Function<Object, Integer>() {
public Integer apply(Object in) {
int index = KNOWN_INDEXES.indexOf(in);
return index == -1 ? null : index;
}
};
...
List<String> values = asList("FOO", "BAR", "PREFA", "ZOO", "PREFZ", "PREFOO");
Collections.sort(values,
Ordering.natural().nullsLast().onResultOf(POSITION_IN_KNOWN_INDEXES).compound(Ordering.natural())
);
So, in other words, sort on natural order of the Integer returned by List.indexOf(), then break ties with natural order of the object itself.
Messy, perhaps, but fun.

I would also use Collections.sort(list) but I think I would use a Comparator and within the comparator you could define your own rules, e.g.
class MyComparator implements Comparator<String> {
public int compare(String o1, String o2) {
// Now you can define the behaviour for your sorting.
// For example your special cases should always come first,
// but if it is not a special case then just use the normal string comparison.
if (o1.equals(SPECIAL_CASE)) {
// Do something special
}
// etc.
return o1.compareTo(o2);
}
}
Then sort by doing:
Collections.sort(list, new MyComparator());

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

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?

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.)

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.
}

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;
}

Four Different Ways to Check If an Array Contains a Value
Using List:
public static boolean useList(String[] arr, String targetValue) {
return Arrays.asList(arr).contains(targetValue);
}
Using Set:
public static boolean useSet(String[] arr, String targetValue) {
Set<String> set = new HashSet<String>(Arrays.asList(arr));
return set.contains(targetValue);
}
Using a simple loop:
public static boolean useLoop(String[] arr, String targetValue) {
for (String s: arr) {
if (s.equals(targetValue))
return true;
}
return false;
}
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

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");

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);
}

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;
}

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:
HashSet
asList
sort & binary
On a sorted array:
HashSet
Binary
asList
So either way, HashSet for the win.

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.

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");

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);

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");
}
}
}

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

the shortest solution
the array VALUES may contain duplicates
since Java 9
List.of(VALUES).contains(s);

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.

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(...)
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);

If you don't want it to be case sensitive
Arrays.stream(VALUES).anyMatch(s::equalsIgnoreCase);

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");
}

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
}
}

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.
}

Use below -
String[] values = {"AB","BC","CD","AE"};
String s = "A";
boolean contains = Arrays.stream(values).anyMatch(v -> v.contains(s));

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

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));
}

Java sort String array of file names by their extension

I have an array of filenames and need to sort that array by the extensions of the filename. Is there an easy way to do this?

Arrays.sort(filenames, new Comparator<String>() {
#Override
public int compare(String s1, String s2) {
// the +1 is to avoid including the '.' in the extension and to avoid exceptions
// EDIT:
// We first need to make sure that either both files or neither file
// has an extension (otherwise we'll end up comparing the extension of one
// to the start of the other, or else throwing an exception)
final int s1Dot = s1.lastIndexOf('.');
final int s2Dot = s2.lastIndexOf('.');
if ((s1Dot == -1) == (s2Dot == -1)) { // both or neither
s1 = s1.substring(s1Dot + 1);
s2 = s2.substring(s2Dot + 1);
return s1.compareTo(s2);
} else if (s1Dot == -1) { // only s2 has an extension, so s1 goes first
return -1;
} else { // only s1 has an extension, so s1 goes second
return 1;
}
}
});
For completeness: java.util.Arrays and java.util.Comparator.

If I remember correctly, the Arrays.sort(...) takes a Comparator<> that it will use to do the sorting. You can provide an implementation of it that looks at the extension part of the string.

You can implement a custom Comparator of Strings. Make it sort them by the substring after the last index of '.'. Then pass in the comparator and your array into
Arrays.sort(stringArray, yourComparator);
// An implementation of the compare method
public int compare(String o1, String o2) {
return o1.substring(o1.lastIndexOf('.')).compareTo(o2.substring(o2.lastIndexOf('.'));
}

Comparators are often hard to get exactly right, and the comparison key has to be generated for every comparison which for most sorting algorithms mean O(n log n). Another approach is to create (key, value) pairs for each item you need to sort, put them in a TreeMap, and then ask for the values as these are sorted according to the key.
For instance
import java.util.Arrays;
import java.util.TreeMap;
public class Bar {
public static void main(String[] args) {
TreeMap<String, String> m2 = new TreeMap<String, String>();
for (String string : Arrays.asList(new String[] { "#3", "#2", "#1" })) {
String key = string.substring(string.length() - 1);
String value = string;
m2.put(key, value);
}
System.out.println(m2.values());
}
}
prints out
[#1, #2, #3]
You should easily be able to adapt the key calculation to your problem.
This only calculates the key once per entry, hence O(n) - (but the sort is still O(n log n)). If the key calculation is expensive or n is large this might be quite measurable.

Create a Comparator and compare the string extensions. Take a look at the following
http://java.sun.com/j2se/1.4.2/docs/api/java/util/Comparator.html
Then pass in your List of strings to Arrays.sort(List, Comparator)

Create your own Comparator that treats the strings as filenames and compares them based on the extensions. Then use Arrays.sort with the Comparator argument.

String DELIMETER = File.separator + ".";
List<String> orginalList = new CopyOnWriteArrayList<>(Arrays.asList(listOfFileNames));
Set<String> setOfuniqueExtension = new TreeSet<>();
for (String item : listOfFileNames) {
if (item.contains(".")) {
String[] split = item.split(DELIMETER);
String temp = "." + split[split.length - 1];
setOfuniqueExtension.add(temp);
}
}
List<String> finalListOfAllFiles = new LinkedList<>();
setOfuniqueExtension.stream().forEach((s1) -> {
for (int i = 0; i < orginalList.size(); i++) {
if (orginalList.get(i).contains(s1)) {
finalListOfAllFiles.add(orginalList.get(i));
orginalList.remove(orginalList.get(i));
i--;
}
}
});
orginalList.stream().filter((s1) -> (!finalListOfAllFiles.contains(s1))).forEach((s1) -> {
finalListOfAllFiles.add(s1);
});
return finalListOfAllFiles;

If you just want to group the files by their extension and do not care about the actual alphabetical order, you can use this:
I think the simplest thing you can do that also works when the filenname does not have a "." is to just reverse the names and compare them.
Arrays.sort(ary, new Comparator<String>() {
#Override
public int compare(String o1, String o2) {
String r1 = new StringBuffer(o1).reverse().toString();
String r2 = new StringBuffer(o2).reverse().toString();
return r1.compareTo(r2);
}
});
Its a shame that java's string does not even have a reverse().