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std::is_permutation

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Sorting operations (on sorted ranges)
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Defined in header <algorithm>
template< class ForwardIt1, class ForwardIt2 >

bool is_permutation( ForwardIt1 first, ForwardIt1 last,

                     ForwardIt2 d_first );
(1) (seit C++11)
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >

bool is_permutation( ForwardIt1 first, ForwardIt1 last,

                     ForwardIt2 d_first, BinaryPredicate p );
(2) (seit C++11)
true kehrt, wenn es eine Permutation der Elemente in dem Bereich, der [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
Original:
Returns true if there exists a permutation of the elements in the range [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
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d_first -
der Beginn des zweiten Bereichs zu vergleichen
Original:
the beginning of the second range to compare
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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.
The types  Type1 and  Type2 must be such that objects of types ForwardIt1 and ForwardIt2 can be dereferenced and then implicitly converted to  Type1 and  Type2 respectively.

Type requirements
-
ForwardIt1, ForwardIt2 must meet the requirements of ForwardIterator.

[Bearbeiten] Rückgabewert

true wenn der Bereich [first, last) ist eine Permutation der Reihe beginnend bei d_first .
Original:
true if the range [first, last) is a permutation of the range beginning at d_first.
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[Bearbeiten] Komplexität

An den meisten O(N2) Anwendungen des Prädikats, oder genau N wenn die Sequenzen schon gleich, wo N=std::distance(first, last) .
Original:
At most O(N2) applications of the predicate, or exactly N if the sequences are already equal, where N=std::distance(first, last).
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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) [edit]
generates the next smaller lexicographic permutation of a range of elements
(Funktions-Template) [edit]