std:: prev_permutation

Transforms the range [ first , last ) into the previous permutation . Returns true if such permutation exists, otherwise transforms the range into the last permutation (as if by std::sort followed by std::reverse ) 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 precedes the old in lexicographical order. false if the first permutation was reached and the range was reset to the last permutation.

Exceptions

Any exceptions thrown from iterator operations or the element swap.

Complexity

Given \(\scriptsize N\) N as std:: distance ( first, last ) :

Possible implementation

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;
        }
    }
}

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 = "cab";
 
    do
    {
        std::cout << s << ' ';
    }
    while (std::prev_permutation(s.begin(), s.end()));
 
    std::cout << s << '\n';
}

cab bca bac acb abc cba

See also