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up vote 1 down vote favorite

The program sorts the array from lowest to highest and outputs the index of the searched value:

public static void main(String[] args) {

    Integer[] random = {6, -4, 12, 0, -10};
    List<Integer> list = new ArrayList<Integer>(Arrays.asList(random));
    Collections.sort(list);
    Integer[] array = list.toArray(new Integer[list.size()]);

    int y = Arrays.binarySearch(array, 6);
    System.out.println(y);

}

This works as expected, but you can see that I had to use 3 lines just to sort the array, before being able to do a binarySearch on it.

Can this code be improved?

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2 Answers 2

up vote 11 down vote accepted

As you say, this is wrong. This

Integer[] random = {6, -4, 12, 0, -10};
List<Integer> list = new ArrayList<Integer>(Arrays.asList(random));
Collections.sort(list);
Integer[] array = list.toArray(new Integer[list.size()]);

could be replaced by

Integer[] random = {6, -4, 12, 0, -10};
List<Integer> list = Arrays.asList(random);
Collections.sort(random);

where Arrays.asList just wraps the array without making a copy. Even better would be

Integer[] random = {6, -4, 12, 0, -10};
Arrays.sort(random);

Also note that sorting something just in order to be able to use binary search once makes no sense.

Also note that

Integer[] random = {6, -4, 12, 0, -10};

uses objects wrapping the int, while

int[] random = {6, -4, 12, 0, -10};

would be much more efficient.

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up vote 2 down vote

Boxing the ints and creating a List are both superfluous. You can just sort an array of ints.

int[] array = {6, -4, 12, 0, -10};
Arrays.sort(array);
System.out.println(Arrays.binarySearch(array, 6));

If your goal is to obtain the approximate rank of an element, it would be faster just to count the elements that are smaller.

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