问题
ALL,
This question is a continuation of this one. I think that STL misses this functionality, but it just my IMHO.
Now, to the question.
Consider following code:
class Foo
{
public:
Foo();
int paramA, paramB;
std::string name;
};
struct Sorter
{
bool operator()(const Foo &foo1, const Foo &foo2) const
{
switch( paramSorter )
{
case 1:
return foo1.paramA < foo2.paramA;
case 2:
return foo1.paramB < foo2.paramB;
default:
return foo1.name < foo2.name;
}
}
int paramSorter;
};
int main()
{
std::vector<Foo> foo;
Sorter sorter;
sorter.paramSorter = 0;
// fill the vector
std::sort( foo.begin(), foo.end(), sorter );
}
At any given moment of time the vector can be re-sorted. The class also have the getter methods which are used in the sorter structure.
What would be the most efficient way to insert a new element in the vector?
Situation I have is:
I have a grid (spreadsheet), that uses the sorted vector of a class. At any given time the vector can be re-sorted and the grid will display the sorted data accordingly.
Now I will need to insert a new element in the vector/grid. I can insert, then re-sort and then re-display the whole grid, but this is very inefficient especially for the big grid.
Any help would be appreciated.
回答1:
The simple answer to the question:
template< typename T >
typename std::vector<T>::iterator
insert_sorted( std::vector<T> & vec, T const& item )
{
return vec.insert
(
std::upper_bound( vec.begin(), vec.end(), item ),
item
);
}
Version with a predicate.
template< typename T, typename Pred >
typename std::vector<T>::iterator
insert_sorted( std::vector<T> & vec, T const& item, Pred pred )
{
return vec.insert
(
std::upper_bound( vec.begin(), vec.end(), item, pred ),
item
);
}
Where Pred is a strictly-ordered predicate on type T.
For this to work the input vector must already be sorted on this predicate.
The complexity of doing this is O(log N) for the upper_bound search (finding where to insert) but up to O(N) for the insert itself.
For a better complexity you could use std::set<T> if there are not going to be any duplicates or std::multiset<T> if there may be duplicates. These will retain a sorted order for you automatically and you can specify your own predicate on these too.
There are various other things you could do which are more complex, e.g. manage a vector and a set / multiset / sorted vector of newly added items then merge these in when there are enough of them. Any kind of iterating through your collection will need to run through both collections.
Using a second vector has the advantage of keeping your data compact. Here your "newly added" items vector will be relatively small so the insertion time will be O(M) where M is the size of this vector and might be more feasible than the O(N) of inserting in the big vector every time. The merge would be O(N+M) which is better than O(NM) it would be inserting one at a time, so in total it would be O(N+M) + O(M²) to insert M elements then merge.
You would probably keep the insertion vector at its capacity too, so as you grow that you will not be doing any reallocations, just moving of elements.
回答2:
If you need to keep the vector sorted all the time, first you might consider whether using std::set or std::multiset won't simplify your code.
If you really need a sorted vector and want to quickly insert an element into it, but do not want to enforce a sorting criterion to be satisfied all the time, then you can first use std::lower_bound() to find the position in a sorted range where the element should be inserted in logarithmic time, then use the insert() member function of vector to insert the element at that position.
If performance is an issue, consider benchmarking std::list vs std::vector. For small items, std::vector is known to be faster because of a higher cache hit rate, but the insert() operation itself is computationally faster on lists (no need to move elements around).
回答3:
Just a note, you can use upper_bound as well depending on your needs. upper_bound will assure new entries that are equivalent to others will appear at the end of their sequence, lower_bound will assure new entries equivalent to others will appear at the beginning of their sequence. Can be useful for certain implementations (maybe classes that can share a "position" but not all of their details!)
Both will assure you that the vector remains sorted according to < result of elements, although inserting into lower_bound will mean moving more elements.
Example:
insert 7 @ lower_bound of { 5, 7, 7, 9 } => { 5, *7*, 7, 7, 9 }
insert 7 @ upper_bound of { 5, 7, 7, 9 } => { 5, 7, 7, *7*, 9 }
回答4:
Instead of inserting and sorting. You should do a find and then insert
Keep the vector sorted. (sort once). When you have to insert
find the first element that compares as greater to the one you are going to insert.
Do an insert just before that position.
This way the vector stays sorted.
Here is an example of how it goes.
start {} empty vector
insert 1 -> find first greater returns end() = 1 -> insert at 1 -> {1}
insert 5 -> find first greater returns end() = 2 -> insert at 2 -> {1,5}
insert 3 -> find first greater returns 2 -> insert at 2 -> {1,3,5}
insert 4 -> find first greater returns 3 -> insert at 3 -> {1,3,4,5}
回答5:
When you want to switch between sort orders, you can use multiple index datastructures, each of which you keep in sorted order (probably some kind of balanced tree, like std::map, which maps sort-keys to vector-indices, or std::set to store pointers to youre obects - but with different comparison functions).
Here's a library which does this: http://www.boost.org/doc/libs/1_53_0/libs/multi_index/doc/index.html
For every change (insert of new elements or update of keys) you must update all index datastructure, or flag them as invalid.
This works if there are not "too many" sort orders and not "too many" updates of your datastructure. Otherwise - bad luck, you have to re-sort everytime you want to change the order.
In other words: The more indices you need (to speed up lookup operations), the more time you need for update operations. And every index needs memory, of course.
To keep the count of indices small, you could use some query engine which combines the indices of several fields to support more complex sort orders over several fields. Like an SQL query optimizer. But that may be overkill...
Example: If you have two fields, a and b, you can support 4 sort orders:
- a
- b
- first a then b
- first b then a
with 2 indices (3. and 4.). With more fields, the possible combinations of sort orders gets big, fast. But you can still use an index which sorts "almost as you want it" and, during the query, sort the remaining fields you couldn't catch with that index, as needed. For sorted output of the whole data, this doesn't help much. But if you only want to lookup some elements, the first "narrowing down" can help much.
回答6:
Assuming you really want to use a vector, and the sort criterium or keys don't change (so the order of already inserted elements always stays the same): Insert the element at the end, then move it to the front one step at a time, until the preceeding element isn't bigger.
It can't be done faster (regarding asymptotic complexity, or "big O notation"), because you must move all bigger elements. And that's the reason why STL doesn't provide this - because it's inefficient on vectors, and you shouldn't use them if you need it.
Edit: Another assumption: Comparing the elements is not much more expensive than moving them. See comments.
Edit 2: As my first assumption doesn't hold (you want to change the sort criterium), scrap this answer and see my other one: https://stackoverflow.com/a/15843955/1413374
来源：https://stackoverflow.com/questions/15843525/how-do-you-insert-the-value-in-a-sorted-vector