std::next_permutation
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Defined in header <algorithm> | ||
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 o
[first, last) intervalo na permutação seguinte a partir do conjunto de todas as permutações que são lexicographically ordenadas em relação a operator< ou comp. Retorna true se existe tal permutação, caso contrário, transforma a faixa para a primeira permutação (como se por std::sort(first, last)) e retorna 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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Índice |
[editar]Parâmetros
| first, last | - | a gama de elementos para 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 se a permutação novo é lexicograficamente maior do que o antigo. false se a permutação última foi alcançado ea gama foi redefinido para a permutação primeiro.
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]Complexidade
Na maioria dos swaps N/2, onde N =std::distance(first, last).
Original:
At most N/2 swaps, where N =std::distance(first, last).
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[editar]Possível implementação
template<class BidirIt>bool next_permutation(BidirIt first, BidirIt last){if(first == last)returnfalse; BidirIt i = last;if(first ==--i)returnfalse; while(1){ BidirIt i1, i2; i1 = i;if(*--i <*i1){ i2 = last;while(!(*i <*--i2));std::iter_swap(i, i2);std::reverse(i1, last);returntrue;}if(i == first){std::reverse(first, last);returnfalse;}}} |
[editar]Exemplo
O código a seguir imprime todas as três permutações da seqüência "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]Veja também
(C++11) | determines if a sequence is a permutation of another sequence (modelo de função) |
| generates the next smaller lexicographic permutation of a range of elements (modelo de função) | |
