std::prev_permutation (3) - Linux Man Pages

std::prev_permutation: std::prev_permutation

NAME

std::prev_permutation - std::prev_permutation

Synopsis


Defined in header <algorithm>
template< class BidirIt > (until C++20)
bool prev_permutation( BidirIt first, BidirIt last);
template< class BidirIt > (since C++20)
constexpr bool prev_permutation( BidirIt first, BidirIt last); (1)
template< class BidirIt, class Compare > (until C++20)
bool prev_permutation( BidirIt first, BidirIt last, Compare comp); (2)
template< class BidirIt, class Compare > (since C++20)
constexpr bool prev_permutation( BidirIt first, BidirIt last, Compare comp);


Transforms the range [first, last) into the previous permutation from the set of all permutations that are lexicographically ordered with respect to operator< or comp. Returns true if such permutation exists, otherwise transforms the range into the last permutation (as if by std::sort(first, last); std::reverse(first, last);) and returns false.

Parameters


first, last - the range of elements to permute
              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);
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 regardless 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 ValueSwappable and 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.

Exceptions


Any exceptions thrown from iterator operations or the element swap.

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.

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;
          }
      }
  }

Example


The following code prints all six permutations of the string "abc" in reverse order
// Run this code


  #include <algorithm>
  #include <string>
  #include <iostream>
  #include <functional>
  int main()
  {
      std::string s="abc";
      std::sort(s.begin(), s.end(), std::greater<char>());
      do {
          std::cout << s << ' ';
      } while(std::prev_permutation(s.begin(), s.end()));
      std::cout << '\n';
  }

Output:


  cba cab bca bac acb abc

See also


is_permutation determines if a sequence is a permutation of another sequence
                 (function template)
(C++11)
                 generates the next greater lexicographic permutation of a range of elements
next_permutation (function template)