std::is_permutation
Defined in header <algorithm>
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(1) | ||
template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(since C++11) (until C++20) |
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template< class ForwardIt1, class ForwardIt2 > constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(since C++20) | |
(2) | ||
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(since C++11) (until C++20) |
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template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(since C++20) | |
(3) | ||
template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(since C++14) (until C++20) |
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template< class ForwardIt1, class ForwardIt2 > constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(since C++20) | |
(4) | ||
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(since C++14) (until C++20) |
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template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(since C++20) | |
Returns true if there exists a permutation of the elements in the range [first1, last1)
that makes that range equal to the range [first2,last2)
, where last2
denotes first2 + (last1 - first1) if it was not given.
operator==
. The behavior is undefined if it is not an equivalence relation.p
. The behavior is undefined if it is not an equivalence relation.Parameters
first1, last1 | - | the range of elements to compare |
first2, last2 | - | the second range to compare |
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 Type &a, const Type &b);
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Type requirements | ||
-ForwardIt1, ForwardIt2 must meet the requirements of LegacyForwardIterator.
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-ForwardIt1, ForwardIt2 must have the same value type.
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Return value
true if the range [first1, last1)
is a permutation of the range [first2, last2)
.
Complexity
At most O(N2) applications of the predicate, or exactly N if the sequences are already equal, where N is std::distance(first1, last1).
However if ForwardIt1
and ForwardIt2
meet the requirements of LegacyRandomAccessIterator and std::distance(first1, last1) != std::distance(first2, last2) no applications of the predicate are made.
Note
The std::is_permutation can be used in testing, namely to check the correctness of rearranging algorithms (e.g. sorting, shuffling, partitioning). If x
is an original range and y
is a permuted range then std::is_permutation(x, y) == true means that y
consist of "the same" elements, maybe staying at other positions.
Possible implementation
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 = std::next(d_first, std::distance(first, last)); for (ForwardIt1 i = first; i != last; ++i) { if (i != std::find(first, i, *i)) continue; // this *i has been checked auto m = std::count(d_first, d_last, *i); if (m==0 || std::count(i, last, *i) != m) { return false; } } } return true; } |
Example
#include <iostream> #include <algorithm> template<typename Os, typename V> Os& operator<< (Os& os, V const& v) { os << "{ "; for (auto const& e : v) os << e << ' '; return os << "}"; } int main() { static constexpr auto v1 = {1,2,3,4,5}; static constexpr auto v2 = {3,5,4,1,2}; static constexpr auto v3 = {3,5,4,1,1}; std::cout << v2 << " is a permutation of " << v1 << ": " << std::boolalpha << std::is_permutation(v1.begin(), v1.end(), v2.begin()) << '\n' << v3 << " is a permutation of " << v1 << ": " << 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
See also
generates the next greater lexicographic permutation of a range of elements (function template) | |
generates the next smaller lexicographic permutation of a range of elements (function template) | |
(C++20) |
specifies that a relation imposes an equivalence relation (concept) |
(C++20) |
determines if a sequence is a permutation of another sequence (niebloid) |