std::next_permutation
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| Déclaré dans l'en-tête <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) | |
Transforme la gamme
[first, last) dans la prochaine permutation de l'ensemble de toutes les permutations qui sont lexicographiquement ordonnées par rapport à operator< ou comp. true Retours s'il en existe permutation, autrement transforme la plage dans la première permutation (comme par std::sort(first, last)) et retourne 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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Sommaire |
[modifier] Paramètres
| first, last | - | la plage d'éléments à permuter
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.
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[modifier] Retourne la valeur
true si la nouvelle permutation est lexicographiquement supérieur à l'ancien. false si la permutation dernier a été atteint et la gamme a été remis à la première permutation .
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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[modifier] Complexité
À la plupart des swaps N/2, où N = std::distance(first, last) .
Original:
At most N/2 swaps, where N = std::distance(first, last).
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[modifier] Mise en œuvre possible
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; } } } |
[modifier] Exemple
Le code suivant imprime tous les trois permutations de la chaîne "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())); }
Résultat :
aab aba baa
[modifier] Voir aussi
| (C++11) |
determines if a sequence is a permutation of another sequence (fonction générique) |
| lexicographique génère le plus petit côté d'une permutation série d'éléments Original: generates the next smaller lexicographic permutation of a range of elements The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. (fonction générique) | |
