std::find_end
De cppreference.com
<metanoindex/>
<tbody> </tbody>| Definido no cabeçalho <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) | |
Procura a subseqüência última
[s_first, s_last) elementos no [first, last) intervalo. A primeira versão utiliza operator== para comparar os elementos, a segunda versão usa o predicado binário dado 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.
Parâmetros
| first, last | - | a gama de elementos para examinar
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 | - | a gama de elementos para pesquisar
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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Valor de retorno
Iterator para o início da subsequência
[s_first, s_last) último intervalo [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.
Se nenhuma subsequência tal for encontrado,
last é retornada. (até C++11)Original:
If no such subsequence is found,
last is returned. (até 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.
Se
[s_first, s_last) está vazio ou se não subseqüência tal for encontrado, last é devolvido. (desde C++11)Original:
If
[s_first, s_last) is empty or if no such subsequence is found, last is returned. (desde 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.
Complexidade
Será que a maioria das comparações dos
S*(N-S+1) onde S = distance(s_first, s_last) e 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.
Possível implementação
| First version |
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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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Exemplo
O código a seguir usa
find_end() para procurar por duas diferentes seqüências de números .
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";
}
}
Saída:
last subsequence is at: 8
subsequence not found