std::find_end

From cppreference.com
< cpp‎ | algorithm
 
 
 
Defined in header <algorithm>
template< class ForwardIt1, class ForwardIt2 >

ForwardIt1 find_end( ForwardIt1 first, ForwardIt1 last,

                     ForwardIt2 s_first, ForwardIt2 s_last );
(1)
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >

ForwardIt1 find_end( ForwardIt1 first, ForwardIt1 last,

                     ForwardIt2 s_first, ForwardIt2 s_last, BinaryPredicate p );
(2)

Searches for the last subsequence of elements [s_first, s_last) in the range [first, last). The first version uses operator== to compare the elements, the second version uses the given binary predicate p.

Contents

[edit] Parameters

first, last - the range of elements to examine
s_first, s_last - the range of elements to search for
p - binary predicate which returns ​true if the elements should be treated as equal.

The signature of the predicate function should be equivalent to the following:

 bool pred(const Type1 &a, const Type2 &b);

The signature does not need to have const &, but the function must not modify the objects passed to it.
The types Type1 and Type2 must be such that objects of types ForwardIt1 and ForwardIt2 can be dereferenced and then implicitly converted to Type1 and Type2 respectively.

Type requirements
-
ForwardIt1 must meet the requirements of ForwardIterator.
-
ForwardIt2 must meet the requirements of ForwardIterator.

[edit] Return value

Iterator to the beginning of last subsequence [s_first, s_last) in range [first, last).

If no such subsequence is found, last is returned.

(until C++11)

If [s_first, s_last) is empty or if no such subsequence is found, last is returned.

(since C++11)

[edit] Complexity

Does at most S*(N-S+1) comparisons where S = distance(s_first, s_last) and N = distance(first, last).

[edit] Possible implementation

First version
template<class ForwardIt1, class ForwardIt2>
ForwardIt1 find_end(ForwardIt1 first, ForwardIt1 last,
                    ForwardIt2 s_first, ForwardIt2 s_last)
{
    if (s_first == s_last)
        return last;
    ForwardIt1 result = last;
    while (1) {
        ForwardIt1 new_result = std::search(first, last, s_first, s_last);
        if (new_result == last) {
            return result;
        } else {
            result = new_result;
            first = result;
            ++first;
        }
    }
    return result;
}
Second version
template<class ForwardIt1, class ForwardIt2, class BinaryPredicate>
ForwardIt1 find_end(ForwardIt1 first, ForwardIt1 last,
                    ForwardIt2 s_first, ForwardIt2 s_last,
                    BinaryPredicate p)
{
    if (s_first == s_last)
        return last;
    ForwardIt1 result = last;
    while (1) {
        ForwardIt1 new_result = std::search(first, last, s_first, s_last, p);
        if (new_result == last) {
            return result;
        } else {
            result = new_result;
            first = result;
            ++first;
        }
    }
    return result;
}

[edit] Example

The following code uses find_end() to search for two different sequences of numbers.

#include <algorithm>
#include <iostream>
#include <vector>
 
int main()
{
    std::vector<int> v{1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4};
    std::vector<int>::iterator result;
 
    std::vector<int> t1{1, 2, 3};
 
    result = std::find_end(v.begin(), v.end(), t1.begin(), t1.end());
    if (result == v.end()) {
        std::cout << "subsequence not found\n";
    } else {
        std::cout << "last subsequence is at: "
                  << std::distance(v.begin(), result) << "\n";
    }
 
    std::vector<int> t2{4, 5, 6};
    result = std::find_end(v.begin(), v.end(), t2.begin(), t2.end());
    if (result == v.end()) {
        std::cout << "subsequence not found\n";
    } else {
        std::cout << "last subsequence is at: " 
                  << std::distance(v.begin(), result) << "\n";
    }
}

Output:

last subsequence is at: 8
subsequence not found

[edit] See also

searches for a range of elements
(function template)
returns true if one set is a subset of another
(function template)
finds the first two adjacent items that are equal (or satisfy a given predicate)
(function template)
finds the first element satisfying specific criteria
(function template)
searches for any one of a set of elements
(function template)
searches for a number consecutive copies of an element in a range
(function template)