Array resort of an in depth analysis of how to put all the odd Numbers in front of all the even Numbers

Here is an efficient algorithm that can be implemented in order (n) time complexity.
The core idea is: Define two Pointers, one pointer A to scan from front to back and one pointer B to scan from back to back. Pointer A scans to an even number to pause, pointer B scans to an odd number to pause, and then exchanges the two Numbers. After the swap, continue to scan and swap as mentioned above, until pointer A and pointer B coincide to stop.
The Java code of this algorithm is as follows:

package Reorder;
public class Reorder {

 public static void main(String[] args) {
  int[] list = { 1, 2, 3, 4, 5, 7, 9, 11 };
  reorderOddEven(list);
 }
 public static void reorderOddEven(int[] list) {
  int length = list.length;
  for (int i = 0; i < length; i++) {
   System.out.print(list[i] + " ");
  }
  System.out.print("n");
  int begin = 0;
  int end = length - 1;
  while (begin < end) {
   while (begin < end && (list[begin] & 0x1) != 0)
    begin++;
   while (begin < end && (list[end] & 0x1) == 0)
    end--;
   if (begin < end) {
    int temp = list[begin];
    list[begin] = list[end];
    list[end] = temp;
   }
  }
  for (int i = 0; i < length; i++) {
   System.out.print(list[i] + " ");
  }
 }
}