If the sub-sequence always starts with a 1 and ends with a 5, you could do this:
row = [1; 2; 3; 4; 5; 1; 1; 2; 3; 4; 5; 5; 1; 1; 2; 2; 3; 4; 5; 5];
start_val = 1;
end_val = 5;
cycle_end_idx = [0; find(diff(row) == start_val-end_val); numel(row)]
n_cycles = numel(cycle_end_idx)-1;
cycle_length = diff(cycle_end_idx);
M = zeros(n_cycles,max(cycle_length));
for ii = 1:n_cycles
M(ii,1:cycle_length(ii)) = row(cycle_end_idx(ii)+1:cycle_end_idx(ii+1));
end
disp(M);
Ultimately, however, you may find that it's better to store those sub-sequences in a cell array rather than a matrix with extra zeros:
C = cell(1,n_cycles);
for ii = 1:n_cycles
C{ii} = row(cycle_end_idx(ii)+1:cycle_end_idx(ii+1)).';
end
format compact
celldisp(C);
