AFAIK, any programming language can do that!  It's not a Processing exclusive matter. 

It's the opposite I did in this post below (1D to 2D). But you need a 3D to 1D: 
http://forum.processing.org/topic/assistance-needed-converting-a-1d-array-to-a-2d-array

Here's my template solution:

// http://forum.processing.org/topic/convert-a-3d-array-to-1d

final int D1 = 10, D2 = 3, D3 = 5;
final int[][][] arr3d = new int[D1][D2][D3];

final int D = D1*D2*D3;
final int[] arr1d = new int[D];

println(D + "\n");

for (int x=0; x!=D1; ++x)  for (int y=0; y!=D2; ++y)  for (int z=0; z!=D3;) {
  final int idx = x*D2*D3 + y*D3 + z;
  arr1d[idx] = arr3d[x][y][z++];
  print(idx + " - ");
}

final int[] sorted = sort(arr1d);

exit();

yes but how? 

like gotoloop did it

here is a lttle longer version


final int D1 = 10, D2 = 3, D3 = 5;
final int[][][] arr3d = new int[D1][D2][D3];

final int D = D1*D2*D3;
final int[] arr1d = new int[D];

println(D + "\n");

// fill 3D
for (int x=0; x!=D1; ++x) 
  for (int y=0; y!=D2; ++y) 
    for (int z=0; z!=D3;) {
      arr3d[x][y][z]=int(random (55));
      print(arr3d[x][y][z++] + " - ");
    }
println ();
println ();
// copy to 1D
for (int x=0; x!=D1; ++x) 
  for (int y=0; y!=D2; ++y) 
    for (int z=0; z!=D3;) {
      final int idx = x*D2*D3 + y*D3 + z;
      arr1d[idx] = arr3d[x][y][z++];
      print(idx + " - ");
    }
println ();
println ();
// sort
final int[] sorted = sort(arr1d);
// show sorted 1D
for (int x=0; x!=sorted.length; ++x)
  print (    sorted[x]  + " - " );
exit();





150

48 - 15 - 35 - 32 - 21 - 15 - 17 - 10 - 27 - 33 - 46 - 48 - 38 - 13 - 10 - 33 - 43 - 45 - 51 - 36 - 0 - 18 - 52 - 12 - 29 - 54 - 41 - 38 - 27 - 29 - 20 - 9 - 29 - 17 - 5 - 29 - 20 - 47 - 40 - 4 - 3 - 0 - 5 - 32 - 52 - 5 - 32 - 17 - 25 - 7 - 9 - 19 - 30 - 5 - 42 - 40 - 32 - 28 - 54 - 51 - 23 - 40 - 26 - 28 - 42 - 32 - 41 - 9 - 54 - 5 - 46 - 35 - 25 - 8 - 20 - 38 - 18 - 47 - 35 - 10 - 47 - 49 - 23 - 38 - 49 - 40 - 0 - 23 - 19 - 54 - 42 - 10 - 15 - 17 - 44 - 22 - 29 - 15 - 23 - 24 - 12 - 25 - 17 - 54 - 30 - 45 - 41 - 53 - 23 - 12 - 0 - 3 - 25 - 53 - 34 - 53 - 26 - 10 - 24 - 15 - 34 - 38 - 1 - 37 - 46 - 4 - 4 - 14 - 35 - 21 - 35 - 50 - 2 - 24 - 21 - 22 - 2 - 5 - 25 - 18 - 32 - 40 - 20 - 33 - 33 - 33 - 23 - 1 - 53 - 30 -

0 - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41 - 42 - 43 - 44 - 45 - 46 - 47 - 48 - 49 - 50 - 51 - 52 - 53 - 54 - 55 - 56 - 57 - 58 - 59 - 60 - 61 - 62 - 63 - 64 - 65 - 66 - 67 - 68 - 69 - 70 - 71 - 72 - 73 - 74 - 75 - 76 - 77 - 78 - 79 - 80 - 81 - 82 - 83 - 84 - 85 - 86 - 87 - 88 - 89 - 90 - 91 - 92 - 93 - 94 - 95 - 96 - 97 - 98 - 99 - 100 - 101 - 102 - 103 - 104 - 105 - 106 - 107 - 108 - 109 - 110 - 111 - 112 - 113 - 114 - 115 - 116 - 117 - 118 - 119 - 120 - 121 - 122 - 123 - 124 - 125 - 126 - 127 - 128 - 129 - 130 - 131 - 132 - 133 - 134 - 135 - 136 - 137 - 138 - 139 - 140 - 141 - 142 - 143 - 144 - 145 - 146 - 147 - 148 - 149 -

0 - 0 - 0 - 0 - 1 - 1 - 2 - 2 - 3 - 3 - 4 - 4 - 4 - 5 - 5 - 5 - 5 - 5 - 5 - 7 - 8 - 9 - 9 - 9 - 10 - 10 - 10 - 10 - 10 - 12 - 12 - 12 - 13 - 14 - 15 - 15 - 15 - 15 - 15 - 17 - 17 - 17 - 17 - 17 - 18 - 18 - 18 - 19 - 19 - 20 - 20 - 20 - 20 - 21 - 21 - 21 - 22 - 22 - 23 - 23 - 23 - 23 - 23 - 23 - 24 - 24 - 24 - 25 - 25 - 25 - 25 - 25 - 26 - 26 - 27 - 27 - 28 - 28 - 29 - 29 - 29 - 29 - 29 - 30 - 30 - 30 - 32 - 32 - 32 - 32 - 32 - 32 - 33 - 33 - 33 - 33 - 33 - 34 - 34 - 35 - 35 - 35 - 35 - 35 - 36 - 37 - 38 - 38 - 38 - 38 - 38 - 40 - 40 - 40 - 40 - 40 - 41 - 41 - 41 - 42 - 42 - 42 - 43 - 44 - 45 - 45 - 46 - 46 - 46 - 47 - 47 - 47 - 48 - 48 - 49 - 49 - 50 - 51 - 51 - 52 - 52 - 53 - 53 - 53 - 53 - 54 - 54 - 54 - 54 - 54 -


Greetings, Chrisir

If you need an answer, please send me a personal message since this forum doesn't notify.

For completeness sake, line -> final int idx = x*D2*D3 + y*D3 + z; is just visually educative. 
We can safely replace it by a simple 1by1 counter:

for (int idx=0, x=0; x!=D1; ++x)  for (int y=0; y!=D2; ++y)  for (int z=0; z!=D3;) {
  //final int idx = x*D2*D3 + y*D3 + z;
  print(idx + " - ");
  arr1d[idx++] = arr3d[x][y][z++];
}

Also, since we know the resultant sorted array is of the same length of its source,
we can use the constant used to create the later (D) in place of sorted.length: 

println('\n');
final int[] sorted = sort(arr1d);

for (int x=0; x!=D; print(sorted[x++]  + " - "));

exit();

Assuming you want to have regularly sized arrays ..... and don't see anything wrong with the array unrolling.