membership test to find if a number is present in array

Question

given an array of int is there a way to do membership test to see if a particular number is present in the int array. the question was to do find missing number between 1-10 given by user. eg [1,2,3,4,5] if 1 is in array it should give true. I cant seem to post my code here.

import java.util.Scanner; 
import java.util.Arrays; 

public class Q1 { 
  public static void main(String[] args) { 
    Scanner scan = new Scanner(System.in); 
    System.out.println("Enter numbers of element: "); 
    int size = scan.nextInt(); 
    int arry[]=new int[size]; 
    for(int i =0;i<size;i++){ 
      System.out.println("Enter any number betweent 1-10:"); 
      int element = scan.nextInt(); 
      arry[i] = element; 
    } 
    System.out.println(Arrays.toString(arry)); 
    System.out.println(Arrays.asList(arry).contains(1)); 
  } 
}

Does this answer your question? [How do I determine whether an array contains a particular value in Java?](https://stackoverflow.com/questions/1128723/how-do-i-determine-whether-an-array-contains-a-particular-value-in-java) — Turamarth, Nov 05 '21 at 14:05
why can't you post your code? code is just flat text, just like the flat text you already posted. — Stultuske, Nov 05 '21 at 14:06
no it didnot answer my question as even if the value was in array it shows false — Ngawang Tobden, Nov 05 '21 at 14:06
i could not post my code as it showed some red dot error even when syntax was correct — Ngawang Tobden, Nov 05 '21 at 14:08
i printed the array and then did membership test. the number was in array but it still showed error. are there other method to see if int array contains a particular value? — Ngawang Tobden, Nov 05 '21 at 14:09
@Ngawang show the code you've done. the only reason why that solution shouldn't work, is because you use a datatype that didn't implement equals — Stultuske, Nov 05 '21 at 14:12
try it in which side of the code? sorry i am just a starter in java. — Ngawang Tobden, Nov 05 '21 at 14:29
instead of int arry, have an Integer[] arry, and arry[i] = new Integer(i); and contains(new Integer(1)); — Stultuske, Nov 08 '21 at 06:20

score 0 · Answer 1 · answered Nov 05 '21 at 15:20

The problem with this code is that Arrays.asList will create List<int[]> NOT the implied List<Integer>, therefore method .contains(1) is always false.

Simple method should be implemented to check the membership:

static boolean isInArray(int n, int ... arr) {
    for (int x : arr) {
        if (x == n) return true;
    }
    return false;
}

Using Stream API:

static boolean isInArray(int n, int ... arr) {
    return Arrays.stream(arr).anyMatch(x -> x == n);
}

Or the input array should be declared as Integer[] arry, then Arrays.asList provides appropriate List<Integer>, and method contains(1) is applicable to such list.

score 0 · Answer 2 · answered Nov 08 '21 at 13:49

Try streams. I don't think aslist().contains() supports primitives like int. This works though:

import java.util.stream.IntStream;

class Main {  
public static void main(String args[]) { 
    int[] a = {1,2,3,4,5,7,8,9};
    int find_1 = 4;
    int find_2 = 6;
    boolean find_num1 = IntStream.of(a).anyMatch(x -> x == find_1);
    boolean find_num2 = IntStream.of(a).anyMatch(x -> x == find_2);
    System.out.println("First num in array? " + find_num1);
    System.out.println("Second num in array? " + find_num2);
    } 
}

membership test to find if a number is present in array

2 Answers2