recursive program help ?

Hi there

Please code in Java 7 . so I will take it and put it in NetBeans ...
Find the attachment .. I just need help with question no 3 .. .. .. feeling hard to get the logic into implementation to find max and min no of recursions

Thank you very much

OK, I've done something now.

It's difficult to see how you can do this without any input prompts at all but, if you think I've done too much, just take them out:

The program examines each of the 'n' elements of the array in turn.

If it's even it examines the next element

If it's odd it looks through the remaining elements until it finds one

(if any) which is even and swaps the two around.

The minimum number of recursions will be none if all array elements

are odd.

The maximum number of recursions will be n - 2 if at least 6 array

elements are even.

import java.util.Scanner;

public class Problem3

{

private static int start = 0;

public static void main(String[] args)

{

Scanner keyboard = new Scanner(System.in);

System.out.print("How many integers are to be entered : ");

int number = keyboard.nextInt();

System.out.println();

if (number < 1)

{

System.out.println("Must be a positive number");

return;

}

System.out.println("Enter " + number + " numbers");

int[] ia = new int[number];

for(int i = 0; i < number; i++)

{

ia[i] = keyboard.nextInt();

}

System.out.println();

ia = evenBeforeOdd(ia);

for (int i : ia)

{

System.out.print(i + " ");

}

private static int[] evenBeforeOdd(int[] ia)

{

for(int i = start; i < ia.length; i++)

{

if (ia[i] % 2 == 0)

{

if (i > start)

{

int temp = ia[i];

ia[i] = ia[start];

ia[start] = temp;

}

start++;

if (start < ia.length - 1)

{

ia = evenBeforeOdd(ia);

}

break;

}

return ia;

}

4 Replies

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Vulpes12y ago
Accepted answer
OK, I've done something now.

It's difficult to see how you can do this without any input prompts at all but, if you think I've done too much, just take them out:

/*
The program examines each of the 'n' elements of the array in turn.

If it's even it examines the next element

If it's odd it looks through the remaining elements until it finds one
(if any) which is even and swaps the two around.

The minimum number of recursions will be none if all array elements
are odd.

The maximum number of recursions will be n - 2 if at least 6 array
elements are even.
*/

import java.util.Scanner;

public class Problem3
{
private static int start = 0;

public static void main(String[] args)
{
Scanner keyboard = new Scanner(System.in);
System.out.print("How many integers are to be entered : ");
int number = keyboard.nextInt();
System.out.println();

if (number < 1)
{
System.out.println("Must be a positive number");
return;
}

System.out.println("Enter " + number + " numbers");
int[] ia = new int[number];

for(int i = 0; i < number; i++)
{
ia[i] = keyboard.nextInt();
}

System.out.println();

ia = evenBeforeOdd(ia);
for (int i : ia)
{
System.out.print(i + " ");
}
}

private static int[] evenBeforeOdd(int[] ia)
{
for(int i = start; i < ia.length; i++)
{
if (ia[i] % 2 == 0)
{
if (i > start)
{
int temp = ia[i];
ia[i] = ia[start];
ia[start] = temp;
}

start++;

if (start < ia.length - 1)
{
ia = evenBeforeOdd(ia);
}

break;
}
}

return ia;
}
}
SUNIL GUTTA12y ago
Ty i will check the fallowing :) Ty so much for your time
SUNIL GUTTA12y ago
sorry for the trouble .. so sorry ..
Tried to attach the pic directly :( later realized it only works if i rar it .. lol .

Cheers
It been too late friend . Take some rest . we can discuss this question 2mrw also .. Ty
Vulpes12y ago
Can't find the attachment.

Can you upload again please.