std:: is_permutation
Defined in header
<algorithm>
|
||
template
<
class
ForwardIt1,
class
ForwardIt2
>
bool
is_permutation
(
ForwardIt1 first1, ForwardIt1 last1,
|
(1) |
(since C++11)
(constexpr since C++20) |
template
<
class
ForwardIt1,
class
ForwardIt2,
class
BinaryPredicate
>
|
(2) |
(since C++11)
(constexpr since C++20) |
template
<
class
ForwardIt1,
class
ForwardIt2
>
bool
is_permutation
(
ForwardIt1 first1, ForwardIt1 last1,
|
(3) |
(since C++14)
(constexpr since C++20) |
template
<
class
ForwardIt1,
class
ForwardIt2,
class
BinaryPredicate
>
|
(4) |
(since C++14)
(constexpr since C++20) |
Checks whether
[
first1
,
last1
)
is a
permutation
of a range starting from
first2
:
- For overloads (1,2) , the second range has std:: distance ( first1, last1 ) elements.
-
For overloads
(3,4)
, the second range is
[
first2,
last2)
.
If
ForwardIt1
and
ForwardIt2
have different
value types
, the program is ill-formed.
If the comparison function is not an equivalence relation , the behavior is undefined.
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 Type1 & a, const Type2 & b ) ;
While the signature does not need to have
const
&
, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const)
|
Type requirements | ||
-
ForwardIt1, ForwardIt2
must meet the requirements of
LegacyForwardIterator
.
|
Return value
true
if the range
[
first1
,
last1
)
is a permutation of the range
[
first2
,
last2
)
,
false
otherwise.
Complexity
Given N as std:: distance ( first1, last1 ) :
ForwardIt1
and
ForwardIt2
are both
LegacyRandomAccessIterator
, and
last1
-
first1
!
=
last2
-
first2
is
true
, no comparison will be made.
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; } |
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.
Example
#include <algorithm> #include <iostream> template<typename Os, typename V> Os& operator<<(Os& os, const V& v) { os << "{ "; for (const auto& 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::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
(algorithm function object) |