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prev_permutation (3)
  • >> prev_permutation (3) ( Solaris man: Библиотечные вызовы )
  • 
                           Standard C++ Library
                 Copyright 1998, Rogue Wave Software, Inc.
    
    
    NAME
         prev_permutation
    
          - Generates successive permutations of a sequence based  on
         an ordering function.
    
    
    
    SYNOPSIS
         #include <algorithm>
         template <class BidirectionalIterator>
         bool prev_permutation (BidirectionalIterator first,
                               BidirectionalIterator last);
    
         template <class BidirectionalIterator, class Compare>
         bool prev_permutation (BidirectionalIterator first,
                               BidirectionalIterator last,
                               Compare comp);
    
    
    
    DESCRIPTION
         The permutation-generating algorithms (next_permutation  and
         prev_permutation) assume that the set of all permutations of
         the elements in a sequence is lexicographically sorted  with
         respect  to  operator<  or  comp. For example, if a sequence
         includes the integers 1 2 3, that sequence has six  permuta-
         tions.  In order from first to last, they are: 1 2 3, 1 3 2,
         2 1 3, 2 3 1, 3 1 2, and 3 2 1.
    
         The prev_permutation algorithm takes a sequence  defined  by
         the  range [first, last) and transforms it into its previous
         permutation, if possible. If such a permutation does  exist,
         the algorithm completes the transformation and returns true.
         If the permutation does not exist, prev_permutation  returns
         false, and transforms the permutation into its "last" permu-
         tation. prev_permutation does the  transformation  according
         to  the lexicographical ordering defined by either operator<
         (the default used in the first version of the algorithm)  or
         comp  (which  is  user-supplied in the second version of the
         algorithm).
    
         For example, if the sequence defined by [first,  last)  con-
         tains  the  integers  1  2 3 (in that order), there is not a
         "previous permutation."  Therefore, the algorithm transforms
         the  sequence  into its last permutation (3 2 1) and returns
         false.
    
    
    
    COMPLEXITY
         At most (last - first)/2 swaps are performed.
    
    
    
    EXAMPLE
         //
         // permute.cpp
         //
          #include <numeric>    //for accumulate
          #include <vector>        //for vector
          #include <functional> //for less
          #include <iostream>
         using namespace std;
    
         int main()
          {
            //Initialize a vector using an array of ints
           int  a1[] = {0,0,0,0,1,0,0,0,0,0};
           char a2[] = "abcdefghji";
    
            //Create the initial set and copies for permuting
           vector<int>  m1(a1, a1+10);
           vector<int>  prev_m1((size_t)10), next_m1((size_t)10);
           vector<char> m2(a2, a2+10);
           vector<char> prev_m2((size_t)10), next_m2((size_t)10);
    
           copy(m1.begin(), m1.end(), prev_m1.begin());
           copy(m1.begin(), m1.end(), next_m1.begin());
           copy(m2.begin(), m2.end(), prev_m2.begin());
           copy(m2.begin(), m2.end(), next_m2.begin());
    
            //Create permutations
            prev_permutation(prev_m1.begin(),
                            prev_m1.end(),less<int>());
           next_permutation(next_m1.begin(),
                            next_m1.end(),less<int>());
            prev_permutation(prev_m2.begin(),
                            prev_m2.end(),less<int>());
           next_permutation(next_m2.begin(),
                            next_m2.end(),less<int>());
            //Output results
           cout << "Example 1: " << endl << "     ";
           cout << "Original values:      ";
           copy(m1.begin(),m1.end(),
                ostream_iterator<int,char>(cout," "));
    
           cout << endl << "     ";
           cout << "Previous permutation: ";
           copy(prev_m1.begin(),prev_m1.end(),
                ostream_iterator<int,char>(cout," "));
    
           cout << endl<< "     ";
           cout << "Next Permutation:     ";
           copy(next_m1.begin(),next_m1.end(),
                ostream_iterator<int,char>(cout," "));
           cout << endl << endl;
    
           cout << "Example 2: " << endl << "     ";
           cout << "Original values: ";
           copy(m2.begin(),m2.end(),
                ostream_iterator<char,char>(cout," "));
           cout << endl << "     ";
           cout << "Previous Permutation: ";
           copy(prev_m2.begin(),prev_m2.end(),
                ostream_iterator<char,char>(cout," "));
           cout << endl << "     ";
    
           cout << "Next Permutation:     ";
           copy(next_m2.begin(),next_m2.end(),
                ostream_iterator<char,char>(cout," "));
           cout << endl << endl;
    
           return 0;
          }
    
         Program Output
    
    
    
         Example 1:
             Original values:      0 0 0 0 1 0 0 0 0 0
             Previous permutation: 0 0 0 0 0 1 0 0 0 0
             Next Permutation:     0 0 0 1 0 0 0 0 0 0
         Example 2:
             Original values: a b c d e f g h j i
             Previous Permutation: a b c d e f g h i j
             Next Permutation:     a b c d e f g i h j
    
    
    
    WARNINGS
         If your compiler does not support default  template  parame-
         ters,  then you always need to supply the Allocator template
         argument. For instance, you need to write:
    
         vector<int, allocator<int> >
         instead of:
    
         vector<int>
    
         If your compiler does not support namespaces,  then  you  do
         not need the using declaration for std.
    
    
    
    SEE ALSO
         next_permutation
    
    
    
    


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