std::find_end
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<metanoindex/>
<tbody> </tbody>| definiert in Header <algorithm>
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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) | |
Sucht nach dem letzten Teilfolge von Elementen
[s_first, s_last) im Bereich [first, last). Die erste Version verwendet operator== um die Elemente zu vergleichen, verwendet die zweite Version des gegebenen binären Prädikats p . Original:
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. The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Parameter
| first, last | - | das Spektrum der Elemente zu untersuchen
Original: the range of elements to examine The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| s_first, s_last | - | der Bereich der Elemente für
Original: the range of elements to search for The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| 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:
The signature does not need to have |
| Type requirements | ||
-ForwardIt1 must meet the requirements of ForwardIterator.
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-ForwardIt2 must meet the requirements of ForwardIterator.
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Rückgabewert
Iterator an den Anfang des letzten Teilfolge
[s_first, s_last) im Bereich [first, last) .Original:
Iterator to the beginning of last subsequence
[s_first, s_last) in range [first, last).The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Wenn keine solche Subsequenz gefunden wird, wird
last zurückgegeben. (bis C + +11)Original:
If no such subsequence is found,
last is returned. (bis C + +11)The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Wenn
[s_first, s_last) leer oder, wenn keine solche Teilfolge gefunden wird, wird last zurückgegeben. (seit C++11)Original:
If
[s_first, s_last) is empty or if no such subsequence is found, last is returned. (seit C++11)The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Komplexität
Ist in den meisten
S*(N-S+1) Vergleiche, wo S = distance(s_first, s_last) und N = distance(first, last) .Original:
Does at most
S*(N-S+1) comparisons where S = distance(s_first, s_last) and N = distance(first, last).The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Mögliche Implementierung
| 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;
}
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| 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;
}
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Beispiel
Der folgende Code verwendet
find_end() für zwei verschiedene Zahlenfolgen suchen .
Original:
The following code uses
find_end() to search for two different sequences of numbers.
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
#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