std::next_permutation() algorithm
- since C++20
- until C++20
// (1)
template< class BidirIt >
constexpr bool next_permutation( BidirIt first, BidirIt last );
// (2)
template< class BidirIt, class Compare >
constexpr bool next_permutation( BidirIt first, BidirIt last, Compare comp );
// (1)
template< class BidirIt >
bool next_permutation( BidirIt first, BidirIt last );
// (2)
template< class BidirIt, class Compare >
bool next_permutation( BidirIt first, BidirIt last, Compare comp );
Permutes the range [first
; last
) into the next permutation,
where the set of all permutations is ordered lexicographically with respect to operator<
or comp
.
Returns true if such a "next permutation" exists;
otherwise transforms the range into the lexicographically first permutation (as if by std::sort(first, last, comp)
) and returns false
.
Parameters
first last | The range of elements to permute. |
comp | Comparison function object (i.e. an object that satisfies the requirements of The signature of the comparison function should be equivalent to the following:
|
Type requirements
ForwardIt | ValueSwappable LegacyBidirectionalIterator |
Return value
true
if the new permutation is lexicographically greater than the old.
false
if the last permutation was reached and the range was reset to the first permutation.
Complexity
Given N
as std::distance(first, last)
At most N / 2
swaps. Averaged over the entire sequence of permutations, typical implementations use about 3
comparisons and 1.5
swaps per call.
Exceptions
Any exceptions thrown from iterator operations or the element swap.
Possible implementation
Implementations (e.g. MSVC STL may enable vectorization when the iterator type satisfies
LegacyContiguousIterator
and swapping its value type calls neither non-trivial special member function nor ADL-found swap.
next_permutation (1)
template<class BidirIt>
bool next_permutation(BidirIt first, BidirIt last)
{
auto r_first = std::make_reverse_iterator(last);
auto r_last = std::make_reverse_iterator(first);
auto left = std::is_sorted_until(r_first, r_last);
if (left != r_last)
{
auto right = std::upper_bound(r_first, left, *left);
std::iter_swap(left, right);
}
std::reverse(left.base(), last);
return left != r_last;
}
Examples
#include <algorithm>
#include <iostream>
#include <string>
int main()
{
std::string s = "aba";
do std::cout << s << '\n';
while (std::next_permutation(s.begin(), s.end()));
std::cout << s << '\n';
}
aba
baa
aab
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