Java Language Arrays Finding an element in an array
Example
There are many ways find the location of a value in an array. The following example snippets all assume that the array is one of the following:
String[] strings = new String[] { "A", "B", "C" };
int[] ints = new int[] { 1, 2, 3, 4 };
In addition, each one sets index or index2 to either the index of required element, or -1 if the element is not present.
Using Arrays.binarySearch (for sorted arrays only)
int index = Arrays.binarySearch(strings, "A");
int index2 = Arrays.binarySearch(ints, 1);
Using a Arrays.asList (for non-primitive arrays only)
int index = Arrays.asList(strings).indexOf("A");
int index2 = Arrays.asList(ints).indexOf(1); // compilation error
Using a Stream
Java SE 8
int index = IntStream.range(0, strings.length)
.filter(i -> "A".equals(strings[i]))
.findFirst()
.orElse(-1); // If not present, gives us -1.
// Similar for an array of primitives
Linear search using a loop
int index = -1;
for (int i = 0; i < array.length; i++) {
if ("A".equals(array[i])) {
index = i;
break;
}
}
// Similar for an array of primitives
Linear search using 3rd-party libraries such as org.apache.commons
int index = org.apache.commons.lang3.ArrayUtils.contains(strings, "A");
int index2 = org.apache.commons.lang3.ArrayUtils.contains(ints, 1);
Note: Using a direct linear search is more efficient than wrapping in a list.
Testing if an array contains an element
The examples above can be adapted to test if the array contains an element by simply testing to see if the index computed is greater or equal to zero.
Alternatively, there are also some more concise variations:
boolean isPresent = Arrays.asList(strings).contains("A");
Java SE 8
boolean isPresent = Stream<String>.of(strings).anyMatch(x -> "A".equals(x));
boolean isPresent = false;
for (String s : strings) {
if ("A".equals(s)) {
isPresent = true;
break;
}
}
boolean isPresent = org.apache.commons.lang3.ArrayUtils.contains(ints, 4);
