I was interviewed a question about implementing a bi-direction
hashtable, i.e. a hashtable that stores key-value pairs such that you
can find value based on key as well as find key based on value.
In another word, there is no obvious difference between a key and a
value, your hashtable should
I have a graph problem, which is different from the standard salesman
problem. I say it is more difficult.
In a graph, I have many vertexes with different colors. It is more
easier to think of each vertex as a shop selling only one goods and
vertexes with the same color selling the same goods.
am agree with Sticker,
I comes to write exactly what Sticker wrote. ;)
It's O(N^2 / 2) that is 1+2+3+4+5+...+N.
On 9/25/07, Sticker [EMAIL PROTECTED] wrote:
I saw your ideas. Some of them are correct some are not. Here is what
I am thinking.
From the question we know that 1 ≤ M
En, it is the idea. You assume each keyword is a single character and
you use a map to hash key words and their counts. Each time you extend
the range to right hand side, you may increase the counts of some
found key words and each time you shrink the range from the left hand
side, you decrease
I have a rough idea about what you are concerning. Yes, DFS is usually
implemented as recursion and there must be stack overflow problem for
any algorithms using stacks to store search paths. But I never think
about crawling the Web as an application of DFS. But it is, I have to
say.
To me, if
to some Information Retrieval
or Database or data structure textbooks to get the right one suits
you.
To me, it is never easy to build a spider that can crawl the
(nontrivial) Web.
On Oct 1, 12:44 pm, Sticker [EMAIL PROTECTED] wrote:
I have a rough idea about what you are concerning. Yes, DFS
, but
it will be
more complex and not as effecient as the above one in practise
Sticker 写道:
Declare: this question is originally from Google. The
original form
is: given a document, how to find a shortest summary
containing all
Declare: this question is originally from Google. The original form
is: given a document, how to find a shortest summary containing all
the key words. After translated, it will be: given a sequence, how to
find a shortest substring that contains all the items required.
Example: a sequence
Can them be implemented using stacks? Two stacks, one is called ls and
the other is rs.
I agree. Two things are needed, 1st, for each trackable action, a
reverse action should be recorded in order to perform the undo action.
The action itself must be recoreded as well for redo action. 2nd, two
to illustrate well by using word typer as only available
here. I will put the link to a pdf file later, showing how I
accomplish the whole problem.
On Sep 5, 6:13 pm, adak [EMAIL PROTECTED] wrote:
Darn it, I meant to post Sticker shock, but my typo left out the
't'. .
Thanks for the clarification
the number
next to min (which is 2). Scan through 6. New segment is {2,3,4,5,6}.
Extend it to 7 but failed.
3. Scan from 3 to 7. New segment {3,4,5,6,7}. Stopped extending it at
8.
4. Scan through 4 to 8 and so on.
I wonder whether there are any better solution to this?
regards
Sticker
On Sep 5
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