Moving the particular numbers to one side of the Array

There is one interesting programming question asked like -
a) You are given an array containing 1 & 0 only & you need to write the code to move all 1s in 
     start of the array & 0s at the end of the array.

b) You are given an array of numbers & you need to write the code to move all the even                numbers in the start of the array & all the odd numbers at the end of the array.

So both the problems are quite easy to suggest the solution & to tell the approach.
But you will see the 1-2 issues when you implement that for the first time & depending on the twist the interviewer adds to it, makes it tricky.
Below is the code I am giving for the above 2 problems to cover different scenarios & please let me know if it fails for any case or you see any more scenario in the same problems & not getting covered below -


​/**
 * @author nitin
 *
 */
public class Defragmentation {
 public static void main(String[] args) {
  //  int[] array = {1,1,1,0,0,1,0,1};
  //    int[] array = {1,0,1,0,1,0,1,0,1,0,1,0};
  //    int[] array = {0,0,0,0,0,0,0};
  //    int[] array = {1,1,1,1,1,1};
  //    int[] array = {1,1,1,1,1,1,0,0,0,0,0,0,0,0};
//  int[] array = {10,12,9,8,7,90,6};
//      int[] array = {2,4,6,8,10,12};
//      int[] array = {1,3,5,7,9};
//    int[] array = {1,3,5,7,9,2,4,6,8,10,12};
  int[] array = {1,3,5,7,9,12};
  defragEven(array);
  System.out.println(array);
 }
 // Below method moves all 1's in the start & all 0's to the end of the Array.
 // It is like stable sort as it will preserve the order of 0's in the Array,
 // if number of 0's & 1's is equal else one element will out of order.
 // Complexity = O(n)
 public static void defrag(int[] array) {
  int second = 0;
  int len = array.length;
  for(int i = 0; i < len; i++) {
   if(array[i] == 0 && array[second] != 0) {
    second = i;
   }
   if(array[i] == 1 && array[second] == 0) {
    int temp = array[i];
    array[i] = array[second];
    array[second] = temp;
    second++;
   }
  }
 }
 // Below method moves all 1's in the start & all 0's to the end of the Array.
 // But will not preserve the order.
 // Complexity = O(n)
 public static void defrag2(int[] array) {
  int len = array.length;
  int zero = len-1;
  for(int i = 0; i < zero;) {
   if(array[i] == 1)
    i++;
   else if(array[zero] == 0)
    zero--;
   else {
    int temp = array[i];
    array[i] = array[zero];
    array[zero] = temp;
    zero--;
    i++;
   }
  }
 }
 // Below method will move all the even numbers to the start & odd numbers to the end of the Array.
 // While doing this it is keeping the order same irrespective of the number of even & odd numbers.
 // Keeping the order same adds to the complexity.
 // Complexity = O(2n) = O(n)
 public static void defragEven(int[] array) {
  int second = 0;
  int len = array.length;
  for(int i = 0; second < len && i < len;) {
   if(array[i]%2 == 0 && array[second]%2 != 0 && i != len-1) {
    int temp = array[i];
    array[i] = array[second];
    array[second] = temp;
    i++;
    second++;
   }
   else if(array[i]%2 == 0 && i < len-1) {
    i++;
   } else if(array[i]%2 != 0 && i < len-1) {
    if(array[second]%2 == 0)
     second = i;
    i++;
   } else {
    while(i > second && array[second]%2 != 0 && array[i]%2 == 0) {
     int temp = array[i-1];
     array[i-1] = array[i];
     array[i] = temp;
     i--;
    }
    break;
   }
  }
 }
 
 // Below method will move all the even numbers to the start & odd numbers to the end of the Array.
 // It doesn't care for the order, so it is bit simpler.
 // Complexity = O(n)
 public static void defragEven2(int[] array) {
  int len = array.length;
  int odd = len-1;
  for(int i = 0; i < odd;) {
   if(array[i]%2 == 0)
    i++;
   else if(array[odd]%2 != 0)
    odd--;
   else {
    int temp = array[i];
    array[i] = array[odd];
    array[odd] = temp;
    odd--;
    i++;
   }
  }
 }
}