std::is_permutation
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Заголовочный файл <algorithm>
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template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first, ForwardIt1 last, |
(1) | (начиная с C++11) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first, ForwardIt1 last, |
(2) | (начиная с C++11) |
[first1, last1)
, что делает этот диапазон равный диапазоне начиная с d_first
. Первый вариант используется operator==
за равноправие, вторая версия использует бинарный p
предикат[first1, last1)
that makes that range equal to the range beginning at d_first
. The first version uses operator==
for equality, the second version uses the binary predicate p
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Содержание |
[править] Параметры
first, last | - | диапазон элементов для сравнения
Original: the range of elements to compare The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. | |||||||||
d_first | - | В начале второго диапазона для сравнения
Original: the beginning of the second range to compare The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. | |||||||||
p | - | binary predicate which returns true if the elements should be treated as equal. The signature of the predicate 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 | |||||||||||
-ForwardIt1, ForwardIt2 must meet the requirements of ForwardIterator .
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[править] Возвращаемое значение
[first, last)
является перестановкой диапазоне начиная с d_first
.[first, last)
is a permutation of the range beginning at d_first
.You can help to correct and verify the translation. Click here for instructions.
[править] Сложность
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[править] Возможная реализация
template<class ForwardIt1, class ForwardIt2> bool is_permutation(ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first) { // skip common prefix std::tie(first, d_first) = std::mismatch(first, last, d_first); // iterate over the rest, counting how many times each element // from [first, last) appears in [d_first, d_last) if (first != last) { ForwardIt2 d_last = d_first; std::advance(d_last, std::distance(first, last)); for (ForwardIt1 i = first; i != last; ++i) { if (i != std::find(first, i, *i)) continue; // already counted this *i auto m = std::count(d_first, d_last, *i); if (m==0 || std::count(i, last, *i) != m) { return false; } } } return true; } |
[править] Пример
#include <algorithm> #include <vector> #include <iostream> int main() { std::vector<int> v1{1,2,3,4,5}; std::vector<int> v2{3,5,4,1,2}; std::cout << "3,5,4,1,2 is a permutation of 1,2,3,4,5? " << std::boolalpha << std::is_permutation(v1.begin(), v1.end(), v2.begin()) << '\n'; std::vector<int> v3{3,5,4,1,1}; std::cout << "3,5,4,1,1 is a permutation of 1,2,3,4,5? " << std::boolalpha << std::is_permutation(v1.begin(), v1.end(), v3.begin()) << '\n'; }
Вывод:
3,5,4,1,2 is a permutation of 1,2,3,4,5? true 3,5,4,1,1 is a permutation of 1,2,3,4,5? false
[править] См. также
generates the next greater lexicographic permutation of a range of elements (шаблон функции) | |
generates the next smaller lexicographic permutation of a range of elements (шаблон функции) |