hi.i've written a program that loads #'s into an array and sorts them.The end user then inputs a number and the program is supposed to perform a sequential search for the input number and tell wether it was found or not found.I'm hoping someone canhelp me with my sequential search.binary gave me enough trouble.any idea?

>binary gave me enough trouble.
Binary search is leaps and bounds more complex than sequential search. Start at the beginning, and go to the end. If you find a match, stop. What's so hard about that? You probably solved the first half of the problem when you printed the sorted array (ie. start at the beginning and go to the end).

/**
 * Requires j2se 5.0 (version 1.5).  This uses class Scanner, static import,
 * and some methods of class Arrays.
 */

import java.util.Arrays;
import java.util.Scanner;
import static java.lang.Math.*;

public class Prog3c_1
{
    public static void main(String... args)
    {
        Scanner sc = new Scanner(System.in);

        int j, k, number, size;
        boolean found = false;

        System.out.print("Enter the array size: ");
        size = sc.nextInt();

        int[] intArray = new int[size];

        for (j=0; j<size; j++)
            intArray[j] = (int) (random()*100);

        System.out.println("A " + size + " element int array of random "
                         + "ints between 0 and 99 has been loaded.");
        System.out.print("Guess one of the numbers: ");
        number = sc.nextInt();

        k=0;
        while (!found && k<size)
        if (intArray[k]==number)
            found=true;
        else
            k +=1;

        System.out.println();
        if (found)
            System.out.println("Found in position " + k);
        else
            System.out.println("The number is not found");

        System.out.println("\nThe random array for verification is:\n");
        System.out.println(Arrays.toString(intArray));
    }
 }
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