问题
I have a situation, as follows:
- I have n doubly-linked lists
- Each list has a sentinel beginning and end
- The lists all have the same beginning and end node (not required, but for simplicity's sake)
- The lists are homogenous and may share items
I'd like to find a partial ordering of all nodes in all n lists, starting with the beginning node and ending with, well, the end node, such that any node which appears in n-x lists, where x < n, will be sorted with respect to the other nodes in all the lists in which it appears.
Using arrays to provide an example set of lists:
first = [a, b, d, f, h, i];
second = [a, b, c, f, g, i];
third = [a, e, f, g, h, i];
Obviously, one possible answer would be [a, b, c, d, e, f, g, h, i], but another admissible ordering would be [a, b, d, e, c, f, g, h, i].
I know that there is a fast algorithm to do this, does anybody remember how it goes or what it is called? I already have a few slow versions, but I'm certain that somewhere in Knuth there is a far faster one.
(And, before you ask, this is not for homework or Project Euler, and I cannot make this any more concrete. This is the problem.)
Edit: I am relatively sure that the partial ordering is defined only as long as the endpoints are in all of the lists and in the same positions (beginning and end). I would not be against a linear-time search to find those endpoints, and if they can't be found, then an error could be raised there.
回答1:
Looks very similar to Topological sort to me. There's several algorithms to get you a topologically sorted state. The one I particularly like is similar to a breadth first search. Maintain two lists, one of all nodes which have no in-edges, say L (initially just the a node), the other with the partial ordered nodes, F. Now at every step,
pick a node from `L`,
do some operations (explained later),
and move the chosen node to the `F` list.
In the "do some operations step",
choose all successors of the source node which have exactly one in-link add them to L.
Remove the link from the source node to all the successors in the previous step
Now, the list F has all your nodes topologically sorted. I'm sorry about the awful explanation, the wiki link has nice diagrams :)
回答2:
Either I'm misunderstanding, or alienhard's algorithm can fail. (I would add this as a comment, but I'm too new here to comment.)
Consider this example:
first = [a, b, c, d, i];
second = [a, d, e, f, i];
third = [a, f, g, h, i];
Alienhard's algorithm will give: a=0, b=1, c=2, d=3, e=2, f=3, g=2, h=3, i=4.
The algorithm then requires that e preceed d even though e follows d in second.
来源：https://stackoverflow.com/questions/6390205/algorithm-for-quickly-obtaining-a-partial-ordering-over-multiple-linked-lists