It took me a while, but I think I understand what you're talking about.
What kind of a solution are you looking for? There are different kinds of solutions. This is how I would program it, using functional programming. This might be too mathy for some people's tastes:
First, we label each number with the list it came from. I want it to be in a certain format so that I can have a list of pairs where the first part of each pair is a list of labels and the last part of each list is the list of numbers that have those labels:
pairs = Union[Thread[{"r", r}], Thread[{"s", s}], Thread[{"t", t}]];
lst = SortBy[Last]@Map[List, pairs, {2}]
The result looks like this:
{{{"t"}, {0}}, {{"r"}, {1}}, {{"s"}, {2}}, {{"r"}, {3}}, {{"t"}, {4}}, {{"t"}, {5}}, {{"s"}, {7}}, {{"r"}, {8}}, {{"t"}, {9}}}
The idea is to turn that list into this list:
{{"t", "r", "s"}, {0, 1, 2}, {"r", "t"}, {3, 4}, {"t", "s", "r"}, {5, 7, 8}, {{"t"}, {9}}}
We have a list of labels and numbers. And each group of labels only has one of each label. I've written up a solution below and attempted to explain it. The solution is written very functionaly. This might be difficult for someone to use without background in this programming style
If the label of the second element occurs in the list of the labels of the first element, then we don't want to join them. Instead we recurse and run the function on the rest of the list, prepending the first element onto the results:
putTogether[{{l1_, n1_}, {{l2_}, {n2_}}, rest___}] /; MemberQ[l1, l2] := Prepend[putTogether[{{{l2}, {n2}}, rest}], {l1, n1}]
Otherwise (if the label of the second element isn't in the labels for the first element), we join them together and recurse:
putTogether[{{l1_, n1_}, {{l2_}, {n2_}}, rest___}] := putTogether[{{Append[l1, l2], Append[n1, n2]}, rest}]
Finally, if none of these patterns match, the function has completed and should just return what it has been given:
putTogether[else_] := else;
This function does the task requested:
Last /@ putTogether[lst]
{{0, 1, 2}, {3, 4}, {5, 7, 8}, {9}}
