Bill, Neil Gianluca, many thanks for your help on this - I will take the time to understand and learn. Great community! Achim

I am astonished the pattern matching approaches are doing as well as they are

...
tbl={Join[Table[1,{10^6}],{2}]};
AbsoluteTiming[Map[g,tbl];]
AbsoluteTiming[Map[ff,tbl];]
AbsoluteTiming[tbl/.{a___Integer,z:Longest[PatternSequence[
  Repeated[x_Integer,{2,Infinity}]]],b___}:> {a,{z}-Range[Length[{z}]-1,0,-1],b};]

Out[1]= {0.169881,Null}
Out[2]= {0.56507,Null}
Out[3]= {0.366563,Null}

When the original message mentioned "efficiently" I assumed this implied far more repetitions and that the timing for pattern matching would grow quadratically, or even exponentially. Thus I tried to think of any single internal function that would find those repetitions without using pattern matching.

I like Bill's solution using Split to find the repeats-- that's clever. Here is a pattern matching approach

In[1]:= ff[{z : Longest[PatternSequence[Repeated[x_, {2, Infinity}]]],
    y___}] := {Range[x - Length[{z}] + 1, x] /. List -> Sequence, y}

In[2]:= ff[{x_}] := {x}

In[3]:= Map[ff, {{2, 2, 7}, {2, 2, 3, 7}, {3}, {6, 6, 6, 7}}]

Out[3]= {{1, 2, 7}, {1, 2, 3, 7}, {3}, {4, 5, 6, 7}}

I made a function that matches a pattern of a repeated sequence, named z, of repeating elements, named x, and the rest of the elements, named y and then reconstructed your list. If you give it a long list it appears to be faster than using split but I am not sure exactly why.

In[9]:= data = 
  Flatten[Table[{{2, 2, 7}, {2, 2, 3, 7}, {3}, {6, 6, 6, 7}}, 10000], 
   1];

In[10]:= Map[ff, data]; // AbsoluteTiming

Out[10]= {0.144403, Null}

In[11]:= Map[g, data]; // AbsoluteTiming

Out[11]= {0.246992, Null}

Regards,

Neil