std:: next_permutation

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.

Parameters

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 ) :

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 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 .

Example

#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

See also