std::prev_permutation
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Заголовочный файл <algorithm>
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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) | |
Преобразование диапазона
[first, last)
в предыдущем перестановку из множества всех перестановок, которые лексикографически упорядочены по operator<
или comp
. Возврат true если такие перестановки существует, иначе превращает диапазон в последние перестановки (как бы std::sort(first, last); std::reverse(first, last);
) и возвращается 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.
You can help to correct and verify the translation. Click here for instructions.
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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 предшествует старых в лексикографическом порядке. false если первая перестановка была достигнута, и диапазон был сброшен до последней перестановки.
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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[править] Сложность
В большинстве свопы
(last-first)/2
.Original:
At most
(last-first)/2
swaps.The text has been machine-translated via Google Translate.
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[править] Возможная реализация
template<class BidirIt> bool prev_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 (*i1 < *--i) { i2 = last; while (!(*--i2 < *i)) ; 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 шесть перестановок строку "ABC" в обратном порядке
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'; }
Вывод:
cba cab bca bac acb abc
[править] См. также
(C++11) |
determines if a sequence is a permutation of another sequence (шаблон функции) |
prev_permutation |
generates the next smaller lexicographic permutation of a range of elements (шаблон функции) |