rationale for std::lower_bound and std::upper_bound?

问题

STL provides binary search functions std::lower_bound and std::upper_bound, but I tend not to use them because I've been unable to remember what they do, because their contracts seem completely mystifying to me.

Just from looking at the names, I'd guess that "lower_bound" might be short for "last lower bound",
i.e. the last element in the sorted list that is <= the given val (if any).
And similarly I'd guess "upper_bound" might be short for "first upper bound",
i.e. the first element in the sorted list that is >= the given val (if any).

But the documentation says they do something rather different from that-- something that seems to be a mixture of backwards and random, to me. To paraphrase the doc:
- lower_bound finds the first element that's >= val
- upper_bound finds the first element that's > val

So lower_bound doesn't find a lower bound at all; it finds the first upper bound!? And upper_bound finds the first strict upper bound.

Does this make any sense?? How do you remember it?

回答1:

If you have multiple elements in the range [first, last) whose value equals the value val you are searching for, then the range [l, u) where

l = std::lower_bound(first, last, val)
u = std::upper_bound(first, last, val)

is precisely the range of elements equal to val within the range [first, last). So l and u are the "lower bound" and "upper bound" for the equal range. It makes sense if you're accustomed to thinking in terms of half-open intervals.

(Note that std::equal_range will return both the lower and upper bound in a pair, in a single call.)

回答2:

std::lower_bound

Returns an iterator pointing to the first element in the range [first, last) that is not less than (i.e. greater or equal to) value.

std::upper_bound

Returns an iterator pointing to the first element in the range [first, last) that is greater than value.

So by mixing both lower and upper bound you are able to exactly describe where your range begins and where it ends.

Does this make any sense??

Yes.

Example:

imagine vector

std::vector<int> data = { 1, 1, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6 };

auto lower = std::lower_bound(data.begin(), data.end(), 4);

1, 1, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6
                  // ^ lower

auto upper = std::upper_bound(data.begin(), data.end(), 4);

1, 1, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6
                           // ^ upper

std::copy(lower, upper, std::ostream_iterator<int>(std::cout, " "));

prints: 4 4 4

http://en.cppreference.com/w/cpp/algorithm/lower_bound

http://en.cppreference.com/w/cpp/algorithm/upper_bound

回答3:

In this case, I think a picture is worth a thousand words. Let's assume we use them to search for 2 in the following collections. The arrows show what iterators the two would return:

So, if you have more than one object with that value already present in the collection, lower_bound will give you an iterator that refers to the first one of them, and upper_bound will give an iterator that refers to the object immediately after the last one of them.

This (among other things) makes the returned iterators usable as the hint parameter to insert.

Therefore, if you use these as the hint, the item you insert will become the new first item with that value (if you used lower_bound) or last item with that value (if you used upper_bound). If the collection didn't contain an item with that value previously, you'll still get an iterator that can be used as a hint to insert it in the correct position in the collection.

Of course, you can also insert without a hint, but using a hint you get a guarantee that the insertion completes with constant complexity, provided that new item to insert can be inserted immediately before the item pointed to by the iterator (as it will in both these cases).

回答4:

Consider the sequence

1 2 3 4 5 6 6 6 7 8 9

lower bound for 6 is the position of the first 6.

upper bound for 6 is the position of the 7.

these positions serve as common (begin, end) pair designating the run of 6-values.

Example:

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

auto main()
    -> int
{
    vector<int> v = {1, 2, 3, 4, 5, 6, 6, 6, 7, 8, 9};
    auto const pos1 = lower_bound( v.begin(), v.end(), 6 );
    auto const pos2 = upper_bound( v.begin(), v.end(), 6 );
    for( auto it = pos1; it != pos2; ++it )
    {
        cout << *it;
    }
    cout << endl;
}

回答5:

I accepted Brian's answer, but I just realized another helpful way of thinking about it which adds clarity for me, so I'm adding this for reference.

Think of the returned iterator as pointing, not at the element *iter, but just before that element, i.e. between that element and the preceding element in the list if there is one. Thinking about it that way, the contracts of the two functions become symmetric: lower_bound finds the position of the transition from <val to >=val, and upper_bound finds the position of the transition from <=val to >val. Or to put it another way, lower_bound is the beginning of the range of items that compare equal to val (i.e. the range that std::equal_range returns), and upper_bound is the end of them.

I wish they would talk about it like this (or any of the other good answers given) in the docs; that would make it much less mystifying!

回答6:

Both functions are very similar, in that they will find an insertion point in a sorted sequence that will preserve the sort. If there are no existing elements in the sequence that are equal to the search item, they will return the same iterator.

If you're trying to find something in the sequence, use lower_bound - it will point directly to the element if it's found.

If you're inserting into the sequence, use upper_bound - it preserves the original ordering of duplicates.

回答7:

the source code actually has a second explanation which I found very helpful to understand the meaning of the function:

lower_bound: Finds the first position in which [val] could be inserted without changing the ordering.

upper_bound: Finds the last position in which [val] could be inserted without changing the ordering.

this [first, last) forms a range which the val could be inserted but still keep the original ordering of the container

lower_bound return "first" i.e. find the "lower boundary of the range"

upper_bound return "last" i.e. find the "upper boundary of the range"

回答8:

There are already good answers as to what std::lower_bound and std::upper_boundis.

I would like to answer your question 'how to remember them'?

Its easy to understand/remember if we draw an analogy with STL begin() and end() methods of any container. begin() returns the starting iterator to the container while end() returns the iterator which is just outside the container and we all know how useful they are while iterating.

Now, on a sorted container and given value, lower_boundand upper_bound will return range of iterators for that value easy to iterate on (just like begin and end)

Practical usage::

Apart form the above mentioned usage on sorted list to access the range for a given value, one of the better applications of upper_bound is to access data having many-to-one relationship in a map.

For example, consider the following relationship: 1 -> a, 2 -> a, 3 -> a, 4 -> b, 5 -> c, 6 -> c, 7 -> c , 8 -> c, 9 -> c, 10 -> c

The above 10 mappings can be saved in map as follows:

numeric_limits<T>::lowest() : UND
1 : a
4 : b
5 : c
11 : UND

The values can be access by the expression (--map.upper_bound(val))->second.

For values of T ranging from lowest to 0, the expression will return UND. For values of T ranging from 1 to 3, it will return 'a' and so on..

Now, imagine we have 100s of data mapping to one value and 100s of such mappings. This approach reduces the size of the map and thereby making it efficient.

回答9:

Yes. The question absolutely has a point. When someone gave these functions their names they were thinking only of sorted arrays with repeating elements. If you have an array with unique elements, "std::lower_bound()" acts more like a search for an "upper bound" unless it finds the actual element.

So this is what I remember about these functions:

If you are doing a binary search, consider using std::lower_bound(), and read the manual. std::binary_search() is based on it, too.
If you want to find the "place" of a value in a sorted array of unique values, consider std::lower_bound() and read the manual.
If you have an arbitrary task of searching in a sorted array, read the manual for both std::lower_bound() and std::upper_bound().

Failing to read the manual after a month or two since you last used these functions, almost certainly leads to a bug.

回答10:

Imagine what you would do if you want to find the first element equal to val in [first, last). You first exclude from first elements which are strictly smaller than val, then exclude backward from last - 1 those strictly greater than val. Then the remaining range is [lower_bound, upper_bound]