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    <p>Easiest way<br> So what I do is first ask the number. Then save the number in a variable before entering it inside the array. Then I check if that number is already in it wih search. If it is I ask for a new number. If it is not I check if it is between 10 and 100, if it is not I ask for a new. I fit is I enter It inside the array. I have to check where an empty place is because the sort mixes up the array everytime</p> <pre><code>import java.util.Arrays; import java.util.Scanner; public class ArrayTest { static Scanner in = new Scanner(System.in); public static void main(String[] args) { int size = 5; int InpNum[] = new int[size]; for (int i = 0; i &lt; InpNum.length; i++){ System.out.println("Please type a number between 10 and 100: "); int number = in.nextInt(); //while (InpNum[i] &lt;= i){ while (Search(InpNum, number) == true){ System.out.println("ERROR: Please enter a number that is not a duplicate of the other numbers you have entered"); number = in.nextInt(); } while (number &lt; 10 || number &gt; 100){ System.out.println("Error: Please type an integer between 10 and 100: "); number = in.nextInt(); } int counter = 0; for (int j = 0; j &lt; InpNum.length &amp;&amp; counter == 0; j++){ if(InpNum[j] == 0){ InpNum[j] = number; counter++; } } Arrays.sort(InpNum); System.out.println(Arrays.toString(InpNum)); //} } } // I can't seem to implement the method below in a useful manner. public static boolean Search(int InpNum[], int searchedNum) { for (int i = 0; i &lt; InpNum.length; i++){ if (InpNum[i] == searchedNum) { return true; } } return false; } } </code></pre>
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    1. POUsing a single-dimensional array, ask the user for input and display only NON-duplicate values
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    1. USDavid Maes
    1. USDavid Maes
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    1. VO
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      1. VTDownMod
    1. COwhy don't you explain yourself a bit?
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      1. USUmNyobe
    2. CODefinitely a working fix, however I'd generally avoid simply providing the solution for questions tagged as `Homework`.
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      1. USulmangt
    3. COOk I edited beacause I forgot the sorting would mix up adding the number.
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      1. USDavid Maes
 

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