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std::next_permutation(3) C++ Standard Libary std::next_permutation(3) NAME std::next_permutation - std::next_permutation Synopsis Defined in header <algorithm> template< class BidirIt > (1) (const- expr since C++20) bool next_permutation( BidirIt first, BidirIt last ); template< class BidirIt, class Compare > bool next_permutation( BidirIt first, BidirIt last, (2) (const- expr since C++20) Compare comp ); 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. 1) The set of all permutations is ordered lexicographically with re- spect to operator< (until C++20) std::less{} (since C++20). 2) The set of all permutations is ordered lexicographically with re- spect to comp. If the type of *first is not Swappable (until C++11) BidirIt is not ValueSwappable (since C++11), the behavior is undefined. Parameters first, last - the range of elements to permute comparison function object (i.e. an object that sat- isfies 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); comp - 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) Type1 and Type2 re- gardless of value category (thus, Type1& is not allowed , nor is Type1 unless for Type1 a move is equivalent to a copy (since C++11)). The types Type1 and Type2 must be such that an object of type BidirIt can be dereferenced and then implicitly converted to both of them. Type requirements - BidirIt must meet the requirements of 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 \(\scriptsize N\)N as std::distance(first, last): 1,2) At most \(\scriptsize \frac{N}{2}\) N 2 swaps. Exceptions Any exceptions thrown from iterator operations or the element swap. 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; } Notes Averaged over the entire sequence of permutations, typical implemen- tations 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. Example The following code prints all three permutations of the string "aba". // Run this code #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 See also is_permutation determines if a sequence is a permutation of another (C++11) sequence (function template) generates the next smaller lexicographic permutation of a prev_permutation range of elements (function template) ranges::next_permutation generates the next greater lexicographic permutation of a (C++20) range of elements (niebloid) http://cppreference.com 2024.06.10 std::next_permutation(3)
NAME | Synopsis | Parameters | Type requirements | Return value | Complexity | Exceptions | Possible implementation | Notes | Example | Output: | See also
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