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In Java
Write a static method named "evenBeforeOdd" that accepts an array of integers as a parameter
and rearranges its elements so that all even values appear before all odds. For example, if the
following array is passed to your method: int[] numbers = {5, 2, 4, 9, 3, 6, 2, 1, 11, 1, 10, 4, 7,
3}; Then after the method has been called, one acceptable ordering of the elements would be: {4,
2, 4, 10, 2, 6, 3, 1, 11, 1, 9, 5, 7, 3} The exact order of the elements does not matter, so long as
all even values appear before all odd values. For example, the following would also be an
acceptable ordering: {2, 2, 4, 4, 6, 10, 1, 1, 3, 3, 5, 7, 9, 11} Do not make any assumptions about
the length of the array or the range of values it might contain. For example, the array might
contain no even elements or no odd elements. You may assume that the array is not null. You
MAY NOT use any temporary arrays to help you solve this problem. (But you may declare as
many simple variables as you like, such as ints.) You also MAY NOT use any other data
structures such as the ArrayList class . DO NOT use Arrays.sort in your solution. Hint: This is
actually a sorting problem. In BubbleSort you tested pairs of elements and swapped them if the
left element is larger than the right element. In this problem, the swap happens when the left
element is odd and the right element is even. You may use the following code to test your
program: If your method is modeled after a bubble sort, expected output is: [2, 4, 6, 2, 10, 4, 5, 9,
3, 1, 11, 1, 7, 3] import java.util.*; public class EvenBeforeOdd { public static void
main(String[] args) { int[] numbers = {5, 2, 4, 9, 3, 6, 2, 1, 11, 1, 10, 4, 7, 3};
evenBeforeOdd(numbers); System.out.println(Arrays.toString(numbers)); } // *** Your method
code goes here *** } // End of EvenBeforeOdd class
Solution
import java.util.*;
public class EvenBeforeOdd {
public static void main(String[] args) {
int[] numbers = {5, 2, 4, 9, 3, 6, 2, 1, 11, 1, 10, 4, 7, 3};
evenBeforeOdd(numbers);
System.out.println(Arrays.toString(numbers));
}
private static void evenBeforeOdd(int[] numbers) {
int len = numbers.length;
int temp;
for(int i=0;i<len;i++)
{
for(int j=i+1;j<len;j++)
{
if(numbers[i]%2==1 && numbers[j]%2==0)
{
temp = numbers[i];
numbers[i] = numbers[j];
numbers[j] = temp;
}
}
}
// TODO Auto-generated method stub
}
}

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In Java Write a static method named --evenBeforeOdd-- that accepts an.docx

  • 1. In Java Write a static method named "evenBeforeOdd" that accepts an array of integers as a parameter and rearranges its elements so that all even values appear before all odds. For example, if the following array is passed to your method: int[] numbers = {5, 2, 4, 9, 3, 6, 2, 1, 11, 1, 10, 4, 7, 3}; Then after the method has been called, one acceptable ordering of the elements would be: {4, 2, 4, 10, 2, 6, 3, 1, 11, 1, 9, 5, 7, 3} The exact order of the elements does not matter, so long as all even values appear before all odd values. For example, the following would also be an acceptable ordering: {2, 2, 4, 4, 6, 10, 1, 1, 3, 3, 5, 7, 9, 11} Do not make any assumptions about the length of the array or the range of values it might contain. For example, the array might contain no even elements or no odd elements. You may assume that the array is not null. You MAY NOT use any temporary arrays to help you solve this problem. (But you may declare as many simple variables as you like, such as ints.) You also MAY NOT use any other data structures such as the ArrayList class . DO NOT use Arrays.sort in your solution. Hint: This is actually a sorting problem. In BubbleSort you tested pairs of elements and swapped them if the left element is larger than the right element. In this problem, the swap happens when the left element is odd and the right element is even. You may use the following code to test your program: If your method is modeled after a bubble sort, expected output is: [2, 4, 6, 2, 10, 4, 5, 9, 3, 1, 11, 1, 7, 3] import java.util.*; public class EvenBeforeOdd { public static void main(String[] args) { int[] numbers = {5, 2, 4, 9, 3, 6, 2, 1, 11, 1, 10, 4, 7, 3}; evenBeforeOdd(numbers); System.out.println(Arrays.toString(numbers)); } // *** Your method code goes here *** } // End of EvenBeforeOdd class Solution import java.util.*; public class EvenBeforeOdd { public static void main(String[] args) { int[] numbers = {5, 2, 4, 9, 3, 6, 2, 1, 11, 1, 10, 4, 7, 3}; evenBeforeOdd(numbers); System.out.println(Arrays.toString(numbers)); } private static void evenBeforeOdd(int[] numbers) { int len = numbers.length; int temp;
  • 2. for(int i=0;i<len;i++) { for(int j=i+1;j<len;j++) { if(numbers[i]%2==1 && numbers[j]%2==0) { temp = numbers[i]; numbers[i] = numbers[j]; numbers[j] = temp; } } } // TODO Auto-generated method stub } }