Checking if an array contains certain integers

Question

I am making an Array that is 10 integers long, so that each place 0-9 contains a different integer 0-9.

I am having trouble figuring out how to check if the array already contains a certain number, and if so, regenerating a new one.

So far I have:

for (int i = 0; i < perm.length; i++)
{
    int num = (int) (Math.random() * 9); 
    int []

    perm[i] = num;   
}

Possible duplicate of [Java, Simplified check if int array contains int](https://stackoverflow.com/questions/12020361/java-simplified-check-if-int-array-contains-int) — borndeft, Aug 04 '17 at 21:59

score 3 · Accepted Answer · edited May 23 '17 at 10:30

3

Arrays.asList(perm).contains(num)

from How can I test if an array contains a certain value?

for (int i = 0; i < perm.length; i++)

this is not enough to loop like this, if collision happens some slots would have been not initalized.

Overall, for this task you better initialize array with values in order and then shuffle it by using random permutation indexes

edited May 23 '17 at 10:30

Community

1
1

answered Dec 09 '12 at 16:12

Nikolay Kuznetsov

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I am new to java and just started learning about arrays. Could you give me an example of what you mean or explain it? – user1729448 Dec 09 '12 at 16:22
@user1729448, explain about what? Have you followed the provided links? – Nikolay Kuznetsov Dec 09 '12 at 16:23
Then `Arrays.asList(perm).contains(num)` answers you question exactly. – Nikolay Kuznetsov Dec 09 '12 at 16:34
Never mind I understand it. So if the num is on the list it returns true? – user1729448 Dec 09 '12 at 16:36
@user1729448, yes, it should. – Nikolay Kuznetsov Dec 09 '12 at 16:37
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/20837/discussion-between-user1729448-and-nikolay-kuznetsov) – user1729448 Dec 09 '12 at 16:39
6

This is wrong.. Arrays.asList(perm) returns a List. `contains` may not always be true, even if the number is in there. – Engineer2021 Apr 25 '14 at 11:59
2

This is wrong. in contains(num) num should be an object. But it will never be the same object even if that number was found early. Because here for each int value a new Integer object is created and passed into the method. – Samitha Chathuranga May 20 '15 at 18:29
1

This will only work for Integer object. Not with primitive int. – Wai Yan Jan 27 '16 at 09:33
This is wrong...per the above comments from @WaiYan – Tim Boland Jun 09 '17 at 16:39

Samitha Chathuranga · Answer 2 · 2017-12-11T04:37:09.577

0

here is a complete answer

int[] array = {3,9, 6, 5, 5, 5, 9, 10, 6, 9,9};

    SortedSet<Integer> s = new TreeSet();
    int numberToCheck=61;
    for (int i = 0; i < array.length; i++) {
        s.add(array[i]);
    }

    System.out.println("Number contains:"+!(s.add(numberToCheck)));


    System.out.println("Sorted list:");
    System.out.print(s);

edited Dec 11 '17 at 04:37

answered May 20 '15 at 19:15

Samitha Chathuranga

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Why do you need to define "mode" and "maxCount" variabled? – MERT DOĞAN Nov 23 '17 at 07:28
Sorry for the confusion. They are not needed. I edited the answer removing them – Samitha Chathuranga Dec 11 '17 at 04:37

score 0 · Answer 3 · answered Aug 03 '18 at 16:26

Adding both string and integer to check is exist or not.

public class existOrNotInArray {

public static void elementExistOrNot() {
    // To Check Character exist or not
    String array[] = { "AB", "CD", "EF" };

    ArrayList arrayList = new ArrayList<>();

    String stringToCheck = "AB";

    for (int i = 0; i < array.length; i++) {
        arrayList.add(array[i]);
    }

    System.out.println("Character exist : " + arrayList.contains(stringToCheck));

    // To Check number exit or not
    int arrayNum[] = { 1, 2, 3, 4 }; // integer array

    int numberToCheck = 5;

    for (int i = 0; i < arrayNum.length; i++) {
        arrayList.add(arrayNum[i]);
    }
    System.out.println("Number exist :" + arrayList.contains(numberToCheck));
}

public static void main(String[] args) {
    elementExistOrNot();
}
}

Checking if an array contains certain integers

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