检查数组是否包含某些整数

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;   
}

Answer 1

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

Answer 2

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);

Answer 3

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();
}
}