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prev_permutation
PrototypePrev_permutation is an overloaded name; there are actually two prev_permutation functions.template <class BidirectionalIterator> bool prev_permutation(BidirectionalIterator first, BidirectionalIterator last); template <class BidirectionalIterator, class StrictWeakOrdering> bool prev_permutation(BidirectionalIterator first, BidirectionalIterator last, StrictWeakOrdering comp); DescriptionPrev_permutation transforms the range of elements [first, last) into the lexicographically next smaller permutation of the elements. There is a finite number of distinct permutations (at most N! [1], where N is last  first), so, if the permutations are ordered by lexicographical_compare, there is an unambiguous definition of which permutation is lexicographically previous. If such a permutation exists, prev_permutation transforms [first, last) into that permutation and returns true. Otherwise it transforms [first, last) into the lexicographically greatest permutation [2] and returns false.The postcondition is that the new permutation of elements is lexicographically less than the old (as determined by lexicographical_compare) if and only if the return value is true. The two versions of prev_permutation differ in how they define whether one element is less than another. The first version compares objects using operator<, and the second compares objects using a function object comp. DefinitionDefined in the standard header algorithm, and in the nonstandard backwardcompatibility header algo.h.Requirements on typesFor the first version:
Preconditions
ComplexityLinear. At most (last  first) / 2 swaps.Exampleint main() { int A[] = {2, 3, 4, 5, 6, 1}; const int N = sizeof(A) / sizeof(int); cout << "Initially: "; copy(A, A+N, ostream_iterator<int>(cout, " ")); cout << endl; prev_permutation(A, A+N); cout << "After prev_permutation: "; copy(A, A+N, ostream_iterator<int>(cout, " ")); cout << endl; next_permutation(A, A+N); cout << "After next_permutation: "; copy(A, A+N, ostream_iterator<int>(cout, " ")); cout << endl; } Notes[1] If all of the elements in [first, last) are distinct from each other, then there are exactly N! permutations. If some elements are the same as each other, though, then there are fewer. There are, for example, only three (3!/2!) permutations of the elements 1 1 2. [2] Note that the lexicographically greatest permutation is, by definition, sorted in nonascending order. See alsonext_permutation, lexicographical_compare, LessThan Comparable, Strict Weak Ordering, sortCopyright © 1999 Silicon Graphics, Inc. All Rights Reserved. TrademarkInformation
