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.
Arrays.binarySearch (for sorted arrays only)
int index = Arrays.binarySearch(strings, "A");
int index2 = Arrays.binarySearch(ints, 1);Arrays.asList (for non-primitive arrays only)
int index = Arrays.asList(strings).indexOf("A");
int index2 = Arrays.asList(ints).indexOf(1); // compilation errorStream
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 primitivesint index = -1;
for (int i = 0; i < array.length; i++) {
if ("A".equals(array[i])) {
index = i;
break;
}
}
// Similar for an array of primitivesint 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.
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");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);