Confirmed.  As noted in the title, this is a problem only when the last position is a 0.

I would prefer not to call out to a second function (repelems) because there is a fair amount of interpreter overhead in calling a function.

Since methods 1 and 3 both work, the only question is performance.  I made a small benchmark

N = 1e4;

bm = zeros (N, 1);

base = [1:10];
rep  = [9:-1:0];

for i = 1:N
  tic;
  c = repelem (base, rep);
  bm(i) = toc;
endfor

The results are

Patch 1:
------------------------------------------------------------
octave:11> mean (bm)
ans =  0.00017877
octave:12> median (bm)
ans =  0.00016284
octave:13> std (bm)
ans =  0.000072822


Patch 3:
------------------------------------------------------------
octave:5> mean (bm)
ans =  0.00015290
octave:4> median (bm)
ans =  0.00013614
octave:6> std (bm)
ans =  0.000080379

Patch three is the winner (although we're only talking about 20 milliseconds).

I committed the patch to the stable branch here (http://hg.savannah.gnu.org/hgweb/octave/rev/9277b77dd28f).

Marking as fixed and closing report.

For example,

octave:1>  repelem(11:13, [1 3 0])
error: index (3): out of bound 2
error: called from
    repelem>prepareIdx at line 317 column 9
    repelem at line 204 column 12

One possible solution is as follow.
(Besides, note that the comment "idx2 output will be a row vector" is wrong; in fact the output can be either row or column vector.)

--- a/scripts/general/repelem.m
+++ b/scripts/general/repelem.m
@@ -299,7 +299,9 @@

   else
     ## This works for a row or column vector.
-    ## idx2 output will be a row vector.
+
+    ## Handle special case if v ends with 0.
+    v(end+1) = 1;

     ## Get ending position for each element item.
     idx_temp = cumsum (v);
@@ -316,6 +318,9 @@
     ## with prepared index
     idx = (find (v != 0))(cumsum (idx));

+    ## Handle special case if v ends with 0.
+    idx(end) = [];
+
   endif

 endfunction

Another solution is to use built-in function "repelems" directly.

--- a/scripts/general/repelem.m
+++ b/scripts/general/repelem.m
@@ -299,22 +299,10 @@

   else
     ## This works for a row or column vector.
-    ## idx2 output will be a row vector.
+    ## idx output will be a row vector.

-    ## Get ending position for each element item.
-    idx_temp = cumsum (v);
-
-    ## Row vector with enough space for output
-    idx(1:idx_temp(end)) = 0;
-
-    ## Set starting position of each element to 1.
-    idx(idx_temp(1:end-1) + 1) = 1;
-
-    ## Set starting position of each element to 1.
-    idx(1) = 1;
-
-    ## with prepared index
-    idx = (find (v != 0))(cumsum (idx));
+    idx_temp = 1:numel (v);
+    idx = repelems (idx_temp, [idx_temp; v(:).']);

   endif

And below is some proposed test case.

--- a/scripts/general/repelem.m
+++ d/scripts/general/repelem.m
@@ -341,6 +341,7 @@
 %!assert (repelem ([-1 0 1], [1 2 1]), [-1 0 0 1])
 %!assert (repelem ([-1 0 1]', [1 2 1]), [-1; 0; 0; 1])
 %!assert (repelem ([1 2 3 4 5]', [2 1 0 1 2]), [1 1 2 4 5 5]')
+%!assert (repelem ([1 2 3 4 5]', [2 1 2 1 0]), [1 1 2 3 3 4]')

 ## nargin == 3 tests
 %!assert (repelem ([1 0;0 -1], 2, 3),
@@ -384,6 +385,7 @@
 %!               1 1 1 1 1 1 0 0 0 1 1 1
 %!               1 1 1 1 1 1 0 0 0 1 1 1]));
 %!assert (repelem ([1 2 3 4 5], 2,[2 1 2 0 2]), [1 1 2 3 3 5 5;1 1 2 3 3 5 5])
+%!assert (repelem ([1 2 3 4 5], 2,[2 1 2 2 0]), [1 1 2 3 3 4 4;1 1 2 3 3 4 4])
 %
 ## nargin > 3 tests
 %!assert (repelem ([1 0;0 -1], 2, 3, 4), ...
@@ -409,6 +411,7 @@
 %!               -1 -1 0 0 0  1  1  1  1
 %!               -1 -1 0 0 0  1  1  1  1]));
 %!assert (repelem ([1 2 3;4 5 6],[0 2],2,2), repmat([4 4 5 5 6 6],2,1,2))
+%!assert (repelem ([1 2 3;4 5 6],[2 0],2,2), repmat([1 1 2 2 3 3],2,1,2))

 ## test with structures
 %!test