Sorting arrays without the Arrays.sort

> I need to create a program which has two integer

> arrays, one with 12 integers, the other with 17. I

> then need to send these two arrays into a new method

> and return another array made up of the two original

> ones. This new array must be in numerical order but

> I'm not allowed to call sort on this array.

Why not?

> Is there

> a way to do this without having to use conditionals

> to test each of the 29 integers?

Maybe your teacher knows a sorting algorithm... you could ask him.

CeciNEstPasUnProgrammeura at 2007-7-29 11:51:25 >

Ok, so I've gotten this far:

import java.util.*;

public class unit7asmntb

{

public static void main (String [] args)

{

int[] arrayTwelve = {1, 23, 29, 99, 5, 17, 73, 34, 50, 8, 81, 82};

int[] arraySeventeen = {2, 4, 9, 75, 83, 14, 98, 56, 65, 63, 12, 88, 55, 27, 78, 94, 30};

Arrays.sort(arrayTwelve);

Arrays.sort(arraySeventeen);

sorter(arrayTwelve, arraySeventeen);

}

public static void sorter (int[] firstArray, int[] secondArray)

{

int[] finale = new int[firstArray.length + secondArray.length];

int firstNumber = 0;

int secondNumber = 0;

int count = 0;

int display = 0;

while ((firstNumber < firstArray.length)&&(secondNumber < secondArray.length))

{

if (firstArray[firstNumber] < secondArray[secondNumber]) {

finale[count] = firstArray[firstNumber];

firstNumber++;

}

else {

finale[count] = secondArray[secondNumber];

secondNumber++;

}

count++;

}

for (display=0; display<finale.length; display++) {

System.out.print(finale[display] + " ");

}

}

}

but when I run it, the last or highest number always appears as 0, no matter what the number is. Why?>

jazelocka at 2007-7-29 11:51:25 >

while ((firstNumber < firstArray.length)&&(secondNumber < secondArray.length))

This is the cause of your problem. Say you have 2 arrays, size 3 and size 5. Your final array will be size 8 but due to your condition in the while loop the max amount of numbers that will be stored in the array will be 5 (all the numbers in the array of 5 for example). At this point secondNumber will no longer be less than secondArray.length and the loop will stop executing and the last 3 elements will all hold the default values of 0.

Another point if you do manage to get all values inserted into the new array, it wont be sorted as you only do a comparison between two numbers at any one time. one array holds 1 8 3 and second array holds 2 7 4 9 1. Your "sorted" array will hold 1 2 7 4 8 3 9 1.

I suggest you place ALL numbers into you new array and once that has been done, sort it.

floundera at 2007-7-29 11:51:25 >