std::is_permutation
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| Definido no cabeçalho <algorithm>
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| template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first, ForwardIt1 last, |
(1) | (desde C++11) |
| template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first, ForwardIt1 last, |
(2) | (desde C++11) |
Retorna true se existe uma permutação dos elementos na
[first1, last1) faixa que faz com que a gama igual a da gama de d_first. A primeira versão utiliza operator== para a igualdade, a segunda versão utiliza o p predicado binárioOriginal:
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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Índice |
[editar] Parâmetros
| first, last | - | a gama de elementos para comparar
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 | - | o início da segunda gama de comparar
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: 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. |
| Type requirements | ||
-ForwardIt1, ForwardIt2 must meet the requirements of ForwardIterator.
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[editar] Valor de retorno
true se o
[first, last) intervalo é uma permutação de início escala em 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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[editar] Complexidade
Na maioria das aplicações de O(N2) do predicado, ou exatamente N se as seqüências já são iguais, onde 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).
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[editar] Possível implementação
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; } |
[editar] Exemplo
#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'; }
Saída:
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
[editar] Veja também
| generates the next greater lexicographic permutation of a range of elements (modelo de função) | |
| generates the next smaller lexicographic permutation of a range of elements (modelo de função) | |
