Where is Java's Array indexOf?

JavaArraysIndexof

Java Problem Overview


I must be missing something very obvious, but I've searched all over and can't find this method.

Java Solutions


Solution 1 - Java

There are a couple of ways to accomplish this using the http://download.oracle.com/javase/1.5.0/docs/api/java/util/Arrays.html">`Arrays`</a> utility class.

If the array is not sorted and is not an array of primitives:

java.util.Arrays.asList(theArray).indexOf(o)

If the array is primitives and not sorted, one should use a solution offered by one of the other answers such as Kerem Baydoğan's, Andrew McKinlay's or Mishax's. The above code will compile even if theArray is primitive (possibly emitting a warning) but you'll get totally incorrect results nonetheless.

If the array is sorted, you can make use of a binary search for performance:

java.util.Arrays.binarySearch(theArray, o)

Solution 2 - Java

Array has no indexOf() method.

Maybe this Apache Commons Lang ArrayUtils method is what you are looking for

import org.apache.commons.lang3.ArrayUtils;

String[] colours = { "Red", "Orange", "Yellow", "Green" };

int indexOfYellow = ArrayUtils.indexOf(colours, "Yellow");

Solution 3 - Java

For primitives, if you want to avoid boxing, Guava has helpers for primitive arrays e.g. Ints.indexOf(int[] array, int target)

Solution 4 - Java

There is none. Either use a java.util.List*, or you can write your own indexOf():

public static <T> int indexOf(T needle, T[] haystack)
{
	for (int i=0; i<haystack.length; i++)
	{
	    if (haystack[i] != null && haystack[i].equals(needle)
	        || needle == null && haystack[i] == null) return i;
    }
	
	return -1;
}

*you can make one from your array using Arrays#asList()

Solution 5 - Java

Unlike in C# where you have the Array.IndexOf method, and JavaScript where you have the indexOf method, Java's API (the Array and Arrays classes in particular) have no such method.

This method indexOf (together with its complement lastIndexOf) is defined in the java.util.List interface. Note that indexOf and lastIndexOf are not overloaded and only take an Object as a parameter.

If your array is sorted, you are in luck because the Arrays class defines a series of overloads of the binarySearch method that will find the index of the element you are looking for with best possible performance (O(log n) instead of O(n), the latter being what you can expect from a sequential search done by indexOf). There are four considerations:

  1. The array must be sorted either in natural order or in the order of a Comparator that you provide as an argument, or at the very least all elements that are "less than" the key must come before that element in the array and all elements that are "greater than" the key must come after that element in the array;

  2. The test you normally do with indexOf to determine if a key is in the array (verify if the return value is not -1) does not hold with binarySearch. You need to verify that the return value is not less than zero since the value returned will indicate the key is not present but the index at which it would be expected if it did exist;

  3. If your array contains multiple elements that are equal to the key, what you get from binarySearch is undefined; this is different from indexOf that will return the first occurrence and lastIndexOf that will return the last occurrence.

  4. An array of booleans might appear to be sorted if it first contains all falses and then all trues, but this doesn't count. There is no override of the binarySearch method that accepts an array of booleans and you'll have to do something clever there if you want O(log n) performance when detecting where the first true appears in an array, for instance using an array of Booleans and the constants Boolean.FALSE and Boolean.TRUE.

If your array is not sorted and not primitive type, you can use List's indexOf and lastIndexOf methods by invoking the asList method of java.util.Arrays. This method will return an AbstractList interface wrapper around your array. It involves minimal overhead since it does not create a copy of the array. As mentioned, this method is not overloaded so this will only work on arrays of reference types.

If your array is not sorted and the type of the array is primitive, you are out of luck with the Java API. Write your own for loop, or your own static utility method, which will certainly have performance advantages over the asList approach that involves some overhead of an object instantiation. In case you're concerned that writing a brute force for loop that iterates over all of the elements of the array is not an elegant solution, accept that that is exactly what the Java API is doing when you call indexOf. You can make something like this:

public static int indexOfIntArray(int[] array, int key) {
    int returnvalue = -1;
    for (int i = 0; i < array.length; ++i) {
        if (key == array[i]) {
            returnvalue = i;
            break;
        }
    }
    return returnvalue;
}

If you want to avoid writing your own method here, consider using one from a development framework like Guava. There you can find an implementation of indexOf and lastIndexOf.

Solution 6 - Java

Java ArrayList has an indexOf method. Java arrays have no such method.

Solution 7 - Java

I don't recall of a "indexOf" on arrays other than coding it for yourself... though you could probably use one of the many java.util.Arrays#binarySearch(...) methods (see the Arrays javadoc) if your array contains primitive types

Solution 8 - Java

The List interface has an indexOf() method, and you can obtain a List from your array with Array's asList() method. Other than that, Array itself has no such method. It does have a binarySearch() method for sorted arrays.

Solution 9 - Java

Arrays themselves do not have that method. A List, however, does: indexOf

Solution 10 - Java

You're probably thinking of the java.util.ArrayList, not the array.

Solution 11 - Java

There is no direct indexOf function in java arrays.

Solution 12 - Java

Jeffrey Hantin's answer is good but it has some constraints, if its this do this or else to that...

You can write your own extension method and it always works the way you want.

Lists.indexOf(array, x -> item == x); // compare in the way you want

And here is your extension

public final class Lists {
    private Lists() {
    }

    public static <T> int indexOf(T[] array, Predicate<T> predicate) {
        for (int i = 0; i < array.length; i++) {
            if (predicate.test(array[i])) return i;
        }
        return -1;
    }

    public static <T> int indexOf(List<T> list, Predicate<T> predicate) {
        for (int i = 0; i < list.size(); i++) {
            if (predicate.test(list.get(i))) return i;
        }
        return -1;
    }

    public interface Predicate<T> {
        boolean test(T t);
    }
}

Solution 13 - Java

int findIndex(int myElement, int[] someArray){
 int index = 0;
 for(int n: someArray){
   if(myElement == n) return index;
   else index++;
 }
}

Note: you can use this method for arrays of type int, you can also use this algorithm for other types with minor changes

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJamieView Question on Stackoverflow
Solution 1 - JavaJeffrey HantinView Answer on Stackoverflow
Solution 2 - JavaKerem BaydoğanView Answer on Stackoverflow
Solution 3 - JavaAndrew McKinlayView Answer on Stackoverflow
Solution 4 - JavaMatt BallView Answer on Stackoverflow
Solution 5 - JavaMishaxView Answer on Stackoverflow
Solution 6 - JavaJamesView Answer on Stackoverflow
Solution 7 - JavaKellindilView Answer on Stackoverflow
Solution 8 - JavaMike YockeyView Answer on Stackoverflow
Solution 9 - JavaIgorView Answer on Stackoverflow
Solution 10 - JavaduffymoView Answer on Stackoverflow
Solution 11 - JavagamboView Answer on Stackoverflow
Solution 12 - JavaM.kazem AkhgaryView Answer on Stackoverflow
Solution 13 - Javavivek guptaView Answer on Stackoverflow