std::is_permutation
Defined in header <algorithm>
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template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(1) | (since C++11) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(2) | (since C++11) |
template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(3) | (since C++14) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(4) | (since C++14) |
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.
The overloads (1) and (3) use operator==
for equality, whereas the overloads (2) and (4) use the binary predicate p
.
Contents |
[edit] 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 ForwardIterator .
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-ForwardIt1, ForwardIt2 must have the same value type.
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[edit] Return value
true if the range [first1, last1)
is a permutation of the range [first2, last2)
.
[edit] Complexity
At most O(N2) applications of the predicate, or exactly N if the sequences are already equal, where N=std::distance(first1, last1).
However if ForwardIt1
and ForwardIt2
meet the requirements of RandomAccessIterator
and std::distance(first1, last1) != std::distance(first2, last2) no applications of the predicate are made.
[edit] 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 = 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; } |
[edit] Example
#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
[edit] 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) |