inplace_merge

發行項
06/09/2015

合併兩個連續排序範圍內的元素至單一排序的範圍，排序準則可能由二進位述詞指定。

template<class BidirectionalIterator>
   void inplace_merge(
      BidirectionalIterator _First, 
      BidirectionalIterator _Middle,
      BidirectionalIterator _Last
   );
template<class BidirectionalIterator, class Predicate>
   void inplace_merge(
      BidirectionalIterator _First, 
      BidirectionalIterator _Middle,
      BidirectionalIterator _Last,
      Predicate _Comp
   );

參數

_First
解決雙向 Iterator 的第一個項目的位置在第一個或兩個連續排序範圍合併和排序成單一範圍。
_Middle
解決雙向 Iterator 的第一個項目的位置在第二個兩個連續排序範圍內結合和排序成單一範圍。
_Last
解決雙向的 Iterator 將最後一個項目的位置會在第二個兩個連續排序範圍內結合和排序成單一範圍。
_Comp
使用者定義的述詞函式物件，定義在此意義上某個項目大於另一個。如果第一個項目則為時，小於第二個項目和 false 二進位述詞會採用兩個引數，則應該傳回 true 。

備註

參考的排序的連續範圍必須是有效的;所有指標必須 dereferenceable，必要時，在每個序列中，最後一個位置必須取用開頭 (由增量。

其中每個引數都必須具有排序的連續範圍，但對 inplace_merge 演算法的應用程式的前提是符合定序與將演算法使用排序合併的範圍。因為項目的相對順序。每個範圍中，儲存作業不穩定。當有兩個來源範圍時的對等項目，此項目是第一個範圍會從第二個元素之前在此合併的範圍。

在演算法配置記憶體到暫存緩衝區，複雜取決於可用的記憶體。如果有足夠的記憶體可用，最佳案例是線性 (_Last – _First) – 1 比較;如果額外的記憶體不可用，最壞情況是 N 記錄*(N)，其中 N = (_Last – _First*)。

範例

// alg_inplace_merge.cpp
// compile with: /EHsc
#include <vector>
#include <algorithm>
#include <functional>      //For greater<int>( )
#include <iostream>

// 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( )
{
   using namespace std;
   vector <int> v1;
   vector <int>::iterator Iter1, Iter2, Iter3;

   // Constructing vector v1 with default less-than ordering
   int i;
   for ( i = 0 ; i <= 5 ; i++ )
   {
      v1.push_back( i );
   }

   int ii;
   for ( ii =-5 ; ii <= 0 ; ii++ )
   {
      v1.push_back(  ii  );
   }

   cout << "Original vector v1 with subranges sorted by the\n "
        <<  "binary predicate less than is  v1 = ( " ;
   for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
      cout << *Iter1 << " ";
   cout << ")" << endl;
   
   // Constructing vector v2 with ranges sorted by greater
   vector <int> v2 ( v1 );
   vector <int>::iterator break2;
   break2 = find ( v2.begin ( ) , v2.end ( ) , -5 );
   sort ( v2.begin ( ) , break2 , greater<int> ( ) );
   sort ( break2 , v2.end ( ) , greater<int> ( ) );
   cout << "Original vector v2 with subranges sorted by the\n "
        << "binary predicate greater is v2 = ( " ;
   for ( Iter2 = v2.begin( ) ; Iter2 != v2.end( ) ; Iter2++ )
      cout << *Iter2 << " ";
   cout << ")" << endl;

   // Constructing vector v3 with ranges sorted by mod_lesser
   vector <int> v3 ( v1 );
   vector <int>::iterator break3;
   break3 = find ( v3.begin ( ) , v3.end ( ) , -5 );
   sort ( v3.begin ( ) , break3 , mod_lesser );
   sort ( break3 , v3.end ( ) , mod_lesser );
   cout << "Original vector v3 with subranges sorted by the\n "
        << "binary predicate mod_lesser is v3 = ( " ;
   for ( Iter3 = v3.begin( ) ; Iter3 != v3.end( ) ; Iter3++ )
      cout << *Iter3 << " ";
   cout << ")" << endl;

   vector <int>::iterator break1;
   break1 = find (v1.begin ( ) , v1.end ( ) , -5 );
   inplace_merge ( v1.begin( ), break1, v1.end( ) );
   cout << "Merged inplace with default order,\n vector v1mod = ( " ;
   for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
      cout << *Iter1 << " ";
   cout << ")" << endl;

   // To merge inplace in descending order, specify binary 
   // predicate greater<int>( )
   inplace_merge ( v2.begin( ), break2 , v2.end( ) , greater<int>( ) );
   cout << "Merged inplace with binary predicate greater specified,\n "
        << "vector v2mod = ( " ;
   for ( Iter2 = v2.begin( ) ; Iter2 != v2.end( ) ; Iter2++ )
      cout << *Iter2 << " ";
   cout << ")" << endl;

   // Applying a user defined (UD) binary predicate mod_lesser
   inplace_merge ( v3.begin( ), break3, v3.end( ), mod_lesser );
   cout << "Merged inplace with binary predicate mod_lesser specified,\n "
        << "vector v3mod = ( " ; ;
   for ( Iter3 = v3.begin( ) ; Iter3 != v3.end( ) ; Iter3++ )
      cout << *Iter3 << " ";
   cout << ")" << endl;
}