std::next_permutation
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| Defined in header <algorithm>
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| template< class BidirIt > bool next_permutation( BidirIt first, BidirIt last ); |
(1) | |
| template< class BidirIt, class Compare > bool next_permutation( BidirIt first, BidirIt last, Compare comp ); |
(2) | |
Transforma el
[first, last) gama en la siguiente permutación del conjunto de todas las permutaciones que están ordenadas lexicográficamente con respecto a operator< o comp. true Devuelve si existe tal permutación, de lo contrario se transforma en el rango de la permutación (como si por std::sort(first, last)) y vuelve false .Original:
Transforms the range
[first, last) into the next permutation from the set of all permutations that are lexicographically ordered with respect to operator< or comp. Returns true if such permutation exists, otherwise transforms the range into the first permutation (as if by std::sort(first, last)) and returns false.The text has been machine-translated via Google Translate.
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Contenido |
[editar] Parámetros
| first, last | - | el intervalo de elementos de permutar
Original: the range of elements to permute 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:
The signature does not need to have const &, but the function must not modify the objects passed to it. | |||||||||
| Type requirements | |||||||||||
-BidirIt must meet the requirements of ValueSwappable and BidirectionalIterator.
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[editar] Valor de retorno
true si la nueva permutación lexicográfica es mayor que el anterior. false si la última permutación fue alcanzado y el intervalo de puesta a cero de la primera permutación .
Original:
true if the new permutation is lexicographically greater than the old. false if the last permutation was reached and the range was reset to the first permutation.
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[editar] Complejidad
En la mayoría de los swaps N/2, donde N = std::distance(first, last) .
Original:
At most N/2 swaps, where N = std::distance(first, last).
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[editar] Posible implementación
template<class BidirIt> bool next_permutation(BidirIt first, BidirIt last) { if (first == last) return false; BidirIt i = last; if (first == --i) return false; while (1) { BidirIt i1, i2; i1 = i; if (*--i < *i1) { i2 = last; while (!(*i < *--i2)) ; std::iter_swap(i, i2); std::reverse(i1, last); return true; } if (i == first) { std::reverse(first, last); return false; } } } |
[editar] Ejemplo
El siguiente código imprime los tres permutaciones de la cadena "aba"
Original:
The following code prints all three permutations of the string "aba"
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#include <algorithm> #include <string> #include <iostream> int main() { std::string s = "aba"; std::sort(s.begin(), s.end()); do { std::cout << s << '\n'; } while(std::next_permutation(s.begin(), s.end())); }
Output:
aab aba baa
[editar] Ver también
| (C++11) |
determines if a sequence is a permutation of another sequence (función de plantilla) |
| generates the next smaller lexicographic permutation of a range of elements (función de plantilla) | |
