How to count the number of consecutive numbers of the same value in an array

I have an array given as

x = [1 1 1 2 2 1 1 1]

I'd like to know a way I could go through each individual element in the array, and getting a value for how many steps the number stays at the same value. For instance, for this example, the output I would be looking for would be

y = [2 1 0 1 0 2 1 0]

Where the first value of 1 stays constant for another 2 steps, the second stays constant for one more step etc.

Answer 1

[Jos (10584)]

% data
  x = [1 1 1 2 2 1 1 1 3 3 3 3 3 5]
% engine
  i = find(diff(x)) 
  n = [i numel(x)] - [0 i]
  c = arrayfun(@(X) X-1:-1:0, n , 'un',0)
  y = cat(2,c{:})

[Gareth Pritchard]

Brilliant, thanks a lot!

[Emil Jensen]

Briliant, tanjs r rot!

Answer 2

[Andrei Bobrov]

c = [1 1 1 2 2 1 1 1];

v = numel(c):-1:1;
ii = [true,diff(c)~=0];
n = v(ii);
t = [n(2:end)+1,1];
out = v - t(cumsum(ii));

Answer 3

[Roger Stafford]

Here's a slightly different way:

 x = [2 2 5 5 5 6 6 6 6 4 7 2 2 2];

 n = size(x,2);
 f = find([true,diff(x)~=0,true]);
 y = zeros(1,n);
 y(f(1:end-1)) = diff(f);
 y = cumsum(y(1:n))-(1:n);

[Gareth Pritchard]

This also works great, thank you. Is there any way you could explain line by line what it is doing at all?