std::set_intersection
![]() |
This page has been machine-translated from the English version of the wiki using Google Translate.
The translation may contain errors and awkward wording. Hover over text to see the original version. You can help to fix errors and improve the translation. For instructions click here. |
Defined in header <algorithm>
|
||
template< class InputIt1, class InputIt2, class OutputIt > OutputIt set_intersection( InputIt1 first1, InputIt1 last1, |
(1) | |
template< class InputIt1, class InputIt2, class OutputIt, class Compare > |
(2) | |
d_first
compuesto de elementos que se encuentran en ambos rangos ordenados [first1, last1)
y [first2, last2)
. La primera versión espera tanto de entrada va a ser resuelto con operator<, la segunda versión espera que se solucionó con la comparación comp
función dada. Si algún elemento se encuentra m
veces en [first1, last1)
y n
veces en [first2, last2)
, los elementos std::min(m, n) primero se copiarán desde el primer rango al rango de destino. El orden de los elementos equivalentes se conserva. El rango resultante no se puede solapar con cualquiera de los rangos de entrada .d_first
consisting of elements that are found in both sorted ranges [first1, last1)
and [first2, last2)
. The first version expects both input ranges to be sorted with operator<, the second version expects them to be sorted with the given comparison function comp
. If some element is found m
times in [first1, last1)
and n
times in [first2, last2)
, the first std::min(m, n) elements will be copied from the first range to the destination range. The order of equivalent elements is preserved. The resulting range cannot overlap with either of the input ranges.You can help to correct and verify the translation. Click here for instructions.
Contenido |
[editar] Parámetros
first1, last1 | - | la primera gama de elementos a examinar
Original: the first 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. |
first2, last2 | - | la segunda gama de elementos a examinar
Original: the second 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. |
comp | - | comparison function which returns true if the first argument is less than the second. The signature of the comparison function should be equivalent to the following: bool cmp(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. |
Type requirements | ||
-InputIt1 must meet the requirements of InputIterator .
| ||
-InputIt2 must meet the requirements of InputIterator .
| ||
-OutputIt must meet the requirements of OutputIterator .
|
[editar] Valor de retorno
You can help to correct and verify the translation. Click here for instructions.
[editar] Complejidad
You can help to correct and verify the translation. Click here for instructions.
[editar] Posible implementación
First version |
---|
template<class InputIt1, class InputIt2, class OutputIt> OutputIt set_intersection(InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first) { while (first1 != last1 && first2 != last2) { if (*first1 < *first2) { ++first1; } else { if (!(*first2 < *first1)) { *d_first++ = *first1++; } ++first2; } } return d_first; } |
Second version |
template<class InputIt1, class InputIt2, class OutputIt, class Compare> OutputIt set_intersection(InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first, Compare comp) { while (first1 != last1 && first2 != last2) { if (comp(*first1, *first2)) { ++first1; } else { if (!comp(*first2, *first1)) { *d_first++ = *first1++; } ++first2; } } return d_first; } |
[editar] Ejemplo
#include <iostream> #include <vector> #include <algorithm> #include <iterator> int main() { std::vector<int> v1{1,2,3,4,5,6,7,8}; std::vector<int> v2{ 5, 7, 9,10}; std::sort(v1.begin(), v1.end()); std::sort(v2.begin(), v2.end()); std::vector<int> v_intersection; std::set_intersection(v1.begin(), v1.end(), v2.begin(), v2.end(), std::back_inserter(v_intersection)); for(int n : v_intersection) std::cout << n << ' '; }
Output:
5 7
[editar] Ver también
computes the union of two sets (función de plantilla) |