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Thread: Problem with LinearArray

  1. #1
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    Default Problem with LinearArray

    Welcome, people

    I have a problem connected with LinearArray. I can search elements with this method but I have a problem with repeated elements. For example, I have elements 1 2 3 1. When I make my method it shows: " Your number is in position 0"? I want to do it with all positions

    --- Update ---

    I mean : " Number 1 is in position 0 , 3.


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    Default Re: Problem with LinearArray

    Post your code.

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    Default Re: Problem with LinearArray

    Sounds like your code is stopping as soon as it finds a match. You need to keep searching the entire array.
    Improving the world one idiot at a time!

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    Default Re: Problem with LinearArray

    public class LinearArray {

    private int data[];
    private static Random generator = new Random();

    public LinearArray(int size) {
    data = new int[size];
    for (int i = 0; i < size; i++) {
    data[i] = 10 + generator.nextInt(91);
    }
    }

    public int lineararray(int searchKey) {
    for (int index = 0; index < data.length; index++) {
    if (data[index] == searchKey) {
    return index;
    }
    }
    return -1;
    }

    public String toString() {
    String temporary = "";
    for (int element : data) {
    temporary += element + " ";
    }
    temporary += " ";
    return temporary;
    }
    }

    --- Update ---

    public static void main(String[] args) {

    Scanner input = new Scanner(System.in);

    int searchInt;
    int position;

    LinearArray line = new LinearArray(10);
    System.out.println(line);

    System.out.print("Enter number(-1 to quit): ");
    searchInt = input.nextInt();

    while (searchInt != -1) {
    position = line.lineararray(searchInt);
    if (position == -1) {
    System.out.println("The integer " + searchInt
    + " was not found.\n");
    } else {
    System.out.println("The integer " + searchInt
    + " was found in position " + position + ".\n");
    }
    System.out.print("Enter number(-1 to quit)");
    searchInt = input.nextInt();
    }
    }
    }

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