std::is_permutation
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<tbody> </tbody>| definiert in Header <algorithm>
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template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first ); |
(1) | (seit C++11) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first, BinaryPredicate p ); |
(2) | (seit C++11) |
true kehrt, wenn es eine Permutation der Elemente in dem Bereich, der [first1, last1) dass Bereichs gleich dem Bereich beginnend bei d_first macht. Die erste Version verwendet operator== für Gleichheit, nutzt die zweite Version der binären Prädikats pOriginal:
Returns
true if there exists a permutation of the elements in the range [first1, last1) that makes that range equal to the range beginning at d_first. The first version uses operator== for equality, the second version uses the binary predicate pThe text has been machine-translated via Google Translate.
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Parameter
| first, last | - | der Bereich von Elementen zu vergleichen
Original: the range of elements to compare The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| d_first | - | der Beginn des zweiten Bereichs zu vergleichen
Original: the beginning of the second range to compare 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, ForwardIt2 must meet the requirements of ForwardIterator.
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Rückgabewert
true wenn der Bereich [first, last) ist eine Permutation der Reihe beginnend bei d_first .Original:
true if the range [first, last) is a permutation of the range beginning at d_first.The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
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Komplexität
An den meisten O(N2) Anwendungen des Prädikats, oder genau N wenn die Sequenzen schon gleich, wo
N=std::distance(first, last) .Original:
At most O(N2) applications of the predicate, or exactly N if the sequences are already equal, where
N=std::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.
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Mögliche Implementierung
template<class ForwardIt1, class ForwardIt2>
bool is_permutation(ForwardIt1 first, ForwardIt1 last,
ForwardIt2 d_first)
{
// skip common prefix
std::tie(first, d_first) = std::mismatch(first, last, d_first);
// iterate over the rest, counting how many times each element
// from [first, last) appears in [d_first, d_last)
if (first != last) {
ForwardIt2 d_last = d_first;
std::advance(d_last, std::distance(first, last));
for (ForwardIt1 i = first; i != last; ++i) {
if (i != std::find(first, i, *i)) continue; // already counted this *i
auto m = std::count(d_first, d_last, *i);
if (m==0 || std::count(i, last, *i) != m) {
return false;
}
}
}
return true;
}
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Beispiel
#include <algorithm>
#include <vector>
#include <iostream>
int main()
{
std::vector<int> v1{1,2,3,4,5};
std::vector<int> v2{3,5,4,1,2};
std::cout << "3,5,4,1,2 is a permutation of 1,2,3,4,5? "
<< std::boolalpha
<< std::is_permutation(v1.begin(), v1.end(), v2.begin()) << '\n';
std::vector<int> v3{3,5,4,1,1};
std::cout << "3,5,4,1,1 is a permutation of 1,2,3,4,5? "
<< std::boolalpha
<< std::is_permutation(v1.begin(), v1.end(), v3.begin()) << '\n';
}
Output:
3,5,4,1,2 is a permutation of 1,2,3,4,5? true
3,5,4,1,1 is a permutation of 1,2,3,4,5? false
Siehe auch
| generates the next greater lexicographic permutation of a range of elements (Funktions-Template) | |
| generates the next smaller lexicographic permutation of a range of elements (Funktions-Template) | |