std::is_permutation
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definiert in Header <algorithm>
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template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first, ForwardIt1 last, |
(1) | (seit C++11) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first, ForwardIt1 last, |
(2) | (seit C++11) |
[first1, last1)
dass Bereichs gleich dem Bereich beginnend bei d_first
macht. Die erste Version verwendet operator==
für Gleichheit, nutzt die zweite Version der binären Prädikats p
[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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Inhaltsverzeichnis |
[Bearbeiten] Parameter
first, last | - | der Bereich von Elementen zu vergleichen
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 | - | der Beginn des zweiten Bereichs zu vergleichen
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: bool pred(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 | ||
-ForwardIt1, ForwardIt2 must meet the requirements of ForwardIterator .
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[Bearbeiten] Rückgabewert
[first, last)
ist eine Permutation der Reihe beginnend bei 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.
[Bearbeiten] Komplexität
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[Bearbeiten] Mögliche Implementierung
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; } |
[Bearbeiten] Beispiel
#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'; }
Output:
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
[Bearbeiten] Siehe auch
generates the next greater lexicographic permutation of a range of elements (Funktions-Template) | |
generates the next smaller lexicographic permutation of a range of elements (Funktions-Template) |