std::next_permutation
Defined in header <algorithm>
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template< class BidirIt > bool next_permutation( BidirIt first, BidirIt last ); |
(1) | (constexpr since C++20) |
template< class BidirIt, class Compare > bool next_permutation( BidirIt first, BidirIt last, Compare comp ); |
(2) | (constexpr since C++20) |
Permutes the range [
first,
last)
into the next permutation. Returns true if such a “next permutation” exists; otherwise transforms the range into the lexicographically first permutation (as if by std::sort) and returns false.
If the type of *first is not Swappable(until C++11)BidirIt
is not ValueSwappable(since C++11), the behavior is undefined.
Contents |
[edit] Parameters
first, last | - | the range of elements to permute |
comp | - | comparison function object (i.e. an object that satisfies the requirements of Compare) which returns true if the first argument is less than the second. The signature of the comparison function should be equivalent to the following: bool cmp(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 | ||
-BidirIt must meet the requirements of LegacyBidirectionalIterator.
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[edit] 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.
[edit] Complexity
Given N as std::distance(first, last):
N |
2 |
[edit] Exceptions
Any exceptions thrown from iterator operations or the element swap.
[edit] Possible implementation
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; } |
[edit] Notes
Averaged over the entire sequence of permutations, typical implementations use about 3 comparisons and 1.5 swaps per call.
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
.
[edit] Example
The following code prints all three permutations of the string "aba".
#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'; }
Output:
aba baa aab
[edit] See also
(C++11) |
determines if a sequence is a permutation of another sequence (function template) |
generates the next smaller lexicographic permutation of a range of elements (function template) | |
(C++20) |
generates the next greater lexicographic permutation of a range of elements (algorithm function object) |