std::prev_permutation() algorithm
- od C++20
- do C++20
// (1)
template< class BidirIt >
constexpr bool prev_permutation( BidirIt first, BidirIt last );
// (2)
template< class BidirIt, class Compare >
constexpr bool prev_permutation( BidirIt first, BidirIt last, Compare comp );
// (1)
template< class BidirIt >
bool prev_permutation( BidirIt first, BidirIt last );
// (2)
template< class BidirIt, class Compare >
bool prev_permutation( BidirIt first, BidirIt last, Compare comp );
Permutes the range [first
; last
) into the previous permutation,
where the set of all permutations is ordered lexicographically with respect to operator<
or comp
.
Returns true if such a permutation exists;
otherwise transforms the range into the lexicographically first permutation (as if by std::sort(first, last, comp); std::reverse(first, last)
) 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 precedes the old in lexicographical order.
false
if the first permutation was reached and the range was reset to the last permutation.
Complexity
At most (last - first) / 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.
prev_permutation (1)
template<class BidirIt>
bool prev_permutation(BidirIt first, BidirIt last)
{
if (first == last)
return false;
BidirIt i = last;
if (first == --i)
return false;
while (1)
{
BidirIt i1, i2;
i1 = i;
if (*i1 < *--i)
{
i2 = last;
while (!(*--i2 < *i))
;
std::iter_swap(i, i2);
std::reverse(i1, last);
return true;
}
if (i == first)
{
std::reverse(first, last);
return false;
}
}
}
Examples
#include <algorithm>
#include <iostream>
#include <string>
int main()
{
std::string s = "cab";
do std::cout << s << ' ';
while (std::prev_permutation(s.begin(), s.end()));
std::cout << s << '\n';
}
cab bca bac acb abc cba
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