prev_permutation

重新排列在範圍內的項目，讓原始順序會用於斷字的上一個更大的變更取代，如果有的話，其中感覺上可能會有一個二進位述詞。

template<class BidirectionalIterator>
   bool prev_permutation(
      BidirectionalIterator _First, 
      BidirectionalIterator _Last
   );
template<class BidirectionalIterator, class BinaryPredicate>
   bool prev_permutation(
      BidirectionalIterator _First, 
      BidirectionalIterator _Last,
      BinaryPredicate _Comp
   );

參數

• _First
  指向第一個項目位置的雙向 Iterator 指向要切換的範圍。

• _Last
  指向超過最後一個項目的位置上的雙向 Iterator 指向要切換的範圍。

• _Comp
  定義連續項目會滿足比較準則順序的使用者定義之述詞函式物件。 一個二進位述詞採用兩個引數並傳回 true ，當內容和 false ，則內容。

傳回值

true ，如果 Dictionary 地變更先前存在並取代原始順序這個範圍，否則， false，在排程先轉換為 Dictionary 地最大的變更情形。

備註

參考的範圍必須是有效的，任何指標必須 dereferenceable，並在序列中最後一個位置開始可取得的會增加。

預設二進位述詞比和該範圍內的項目必須小於可比較必定小於之前變更已妥善定義的。

複雜度是線性，與最多 _Last (–) _First/2 交換。

範例

// alg_prev_perm.cpp
// compile with: /EHsc
#include <vector>
#include <deque>
#include <algorithm>
#include <iostream>
#include <ostream>

using namespace std;
class CInt;
ostream& operator<<( ostream& osIn, const CInt& rhs );

class CInt {
public:
   CInt( int n = 0 ) : m_nVal( n ){}
   CInt( const CInt& rhs ) : m_nVal( rhs.m_nVal ){}
   CInt&   operator=( const CInt& rhs ) {m_nVal =
   rhs.m_nVal; return *this;}
   bool operator<( const CInt& rhs ) const
      {return ( m_nVal < rhs.m_nVal );}
   friend ostream& operator<<( ostream& osIn, const CInt& rhs );

private:
   int m_nVal;
};

inline ostream& operator<<( ostream& osIn, const CInt& rhs ) {
   osIn << "CInt( " << rhs.m_nVal << " )";
   return osIn;
}

// Return whether modulus of elem1 is less than modulus of elem2
bool mod_lesser (int elem1, int elem2 ) {
   if ( elem1 < 0 )
      elem1 = - elem1;
   if ( elem2 < 0 )
      elem2 = - elem2;
   return elem1 < elem2;
};

int main() {
   // Reordering the elements of type CInt in a deque
   // using the prev_permutation algorithm
   CInt c1 = 1, c2 = 5, c3 = 10;
   bool deq1Result;
   deque<CInt> deq1, deq2, deq3;
   deque<CInt>::iterator d1_Iter;

   deq1.push_back ( c1 );
   deq1.push_back ( c2 );
   deq1.push_back ( c3 );

   cout << "The original deque of CInts is deq1 = (";
   for ( d1_Iter = deq1.begin( ); d1_Iter != --deq1.end( ); d1_Iter++ )
      cout << " " << *d1_Iter << ",";
   d1_Iter = --deq1.end( );
   cout << " " << *d1_Iter << " )." << endl;

   deq1Result = prev_permutation ( deq1.begin ( ) , deq1.end ( ) );

   if ( deq1Result )
      cout << "The lexicographically previous permutation "
           << "exists and has \nreplaced the original "
           << "ordering of the sequence in deq1." << endl;
   else
      cout << "The lexicographically previous permutation doesn't "
           << "exist\n and the lexicographically "
           << "smallest permutation\n has replaced the "
           << "original ordering of the sequence in deq1." << endl;

   cout << "After one application of prev_permutation,\n deq1 = (";
   for ( d1_Iter = deq1.begin( ); d1_Iter != --deq1.end( ); d1_Iter++ )
      cout << " " << *d1_Iter << ",";
   d1_Iter = --deq1.end( );
   cout << " " << *d1_Iter << " )." << endl << endl;

   // Permutating vector elements with binary function mod_lesser
   vector <int> v1;
   vector <int>::iterator Iter1;

   int i;
   for ( i = -3 ; i <= 3 ; i++ )
      v1.push_back( i );

   cout << "Vector v1 is ( " ;
   for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
      cout << *Iter1 << " ";
   cout << ")." << endl;

   prev_permutation ( v1.begin ( ) , v1.end ( ) , mod_lesser );

   cout << "After the first prev_permutation, vector v1 is:\n v1 = ( " ;
   for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
      cout << *Iter1 << " ";
   cout << ")." << endl;

   int iii = 1;
   while ( iii <= 5 ) {
      prev_permutation ( v1.begin ( ) , v1.end ( ) , mod_lesser );
      cout << "After another prev_permutation of vector v1,\n v1 =   ( " ;
      for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ;Iter1 ++ )
         cout << *Iter1  << " ";
      cout << ")." << endl;
      iii++;
   }
}

需求

標題: <algorithm>

命名空間: std

請參閱

參考

prev_permutation (STL Samples)

標準樣板程式庫