std::next_permutation
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Заголовочный файл <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) | |
Преобразование диапазона
[first, last)
в следующем перестановку из множества всех перестановок, которые лексикографически упорядочены по operator<
или comp
. Возврат true если такие перестановки существует, иначе превращает диапазон в первую перестановку (как бы std::sort(first, last)
) и возвращается 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.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Содержание |
[править] Параметры
first, last | - | диапазон элементов переставлять
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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[править] Возвращаемое значение
true если новые перестановки лексикографически больше, чем старые. false если последняя перестановка была достигнута, и диапазон был сброшен в первую перестановку.
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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[править] Сложность
В большинстве свопы N/2, где N = std::distance(first, last).
Original:
At most N/2 swaps, where N = std::distance(first, last).
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[править] Возможная реализация
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; } } } |
[править] Пример
Следующий код выводит все три перестановки строки "аба"
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())); }
Вывод:
aab aba baa
[править] См. также
(C++11) |
determines if a sequence is a permutation of another sequence (шаблон функции) |
generates the next smaller lexicographic permutation of a range of elements (шаблон функции) |