std::prev_permutation
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Defined in header <algorithm> | ||
template<class BidirIt > bool prev_permutation( BidirIt first, BidirIt last); | (1) | |
template<class BidirIt, class Compare > bool prev_permutation( BidirIt first, BidirIt last, Compare comp); | (2) | |
Transforma el
[first, last) gama en la permutación anterior a partir 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 final (como por std::sort(first, last); std::reverse(first, last);) y vuelve false .Original:
Transforms the range
[first, last) into the previous 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 last permutation (as if by std::sort(first, last); std::reverse(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: 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 | ||
-BidirIt must meet the requirements of ValueSwappable and BidirectionalIterator. | ||
[editar]Valor de retorno
true si la nueva permutación precede a la edad en orden lexicográfico. false si la primera permutación fue alcanzado y el intervalo de puesta a cero a la última permutación .
Original:
true if the new permutation precedes the old in lexicographical order. false if the first permutation was reached and the range was reset to the last permutation.
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[editar]Complejidad
En la mayoría de los swaps
(last-first)/2 .Original:
At most
(last-first)/2 swaps.The text has been machine-translated via Google Translate.
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[editar]Posible implementación
template<class BidirIt>bool prev_permutation(BidirIt first, BidirIt last){if(first == last)returnfalse; BidirIt i = last;if(first ==--i)returnfalse; while(1){ BidirIt i1, i2; i1 = i;if(*i1 <*--i){ i2 = last;while(!(*--i2 <*i));std::iter_swap(i, i2);std::reverse(i1, last);returntrue;}if(i == first){std::reverse(first, last);returnfalse;}}} |
[editar]Ejemplo
El siguiente código imprime los seis permutaciones de la cadena "abc" en orden inverso
Original:
The following code prints all six permutations of the string "abc" in reverse order
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#include <algorithm>#include <string>#include <iostream>#include <functional>int main(){std::string s="abc";std::sort(s.begin(), s.end(), std::greater<char>());do{std::cout<< s <<' ';}while(std::prev_permutation(s.begin(), s.end()));std::cout<<'\n';}
Output:
cba cab bca bac acb abc
[editar]Ver también
(C++11) | determines if a sequence is a permutation of another sequence (función de plantilla) |
prev_permutation | generates the next smaller lexicographic permutation of a range of elements (función de plantilla) |
