C++ Tutorial/STL Algorithms Binary search/upper bound

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lower_bound and upper_bound

/* The following code example is taken from the book
 * "The C++ Standard Library - A Tutorial and Reference"
 * by Nicolai M. Josuttis, Addison-Wesley, 1999
 *
 * (C) Copyright Nicolai M. Josuttis 1999.
 * Permission to copy, use, modify, sell and distribute this software
 * is granted provided this copyright notice appears in all copies.
 * This software is provided "as is" without express or implied
 * warranty, and with no claim as to its suitability for any purpose.
 */
#include <iostream>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <iterator>
#include <functional>
#include <numeric>
using namespace std;
/* PRINT_ELEMENTS()
 * - prints optional C-string optcstr followed by
 * - all elements of the collection coll
 * - separated by spaces
 */
template <class T>
inline void PRINT_ELEMENTS (const T& coll, const char* optcstr="")
{
    typename T::const_iterator pos;
    std::cout << optcstr;
    for (pos=coll.begin(); pos!=coll.end(); ++pos) {
        std::cout << *pos << " ";
    }
    std::cout << std::endl;
}
/* INSERT_ELEMENTS (collection, first, last)
 * - fill values from first to last into the collection
 * - NOTE: NO half-open range
 */
template <class T>
inline void INSERT_ELEMENTS (T& coll, int first, int last)
{
    for (int i=first; i<=last; ++i) {
        coll.insert(coll.end(),i);
    }
}

int main()
{
    list<int> coll;
    INSERT_ELEMENTS(coll,1,9);
    INSERT_ELEMENTS(coll,1,9);
    coll.sort ();
    PRINT_ELEMENTS(coll);
    // print first and last position 5 could get inserted
    list<int>::iterator pos1, pos2;
    pos1 = lower_bound (coll.begin(), coll.end(), 5);
    pos2 = upper_bound (coll.begin(), coll.end(), 5);
    cout << "5 could get position " << distance(coll.begin(),pos1) + 1
         << " up to "               << distance(coll.begin(),pos2) + 1
         << " without breaking the sorting" << endl;
    // insert 3 at the first possible position without breaking the sorting
    coll.insert (lower_bound(coll.begin(),coll.end(),
                             3),
                 3);
    // insert 7 at the last possible position without breaking the sorting
    coll.insert (upper_bound(coll.begin(),coll.end(),
                             7),
                 7);
    PRINT_ELEMENTS(coll);
}
1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9
5 could get position 9 up to 11 without breaking the sorting
1 1 2 2 3 3 3 4 4 5 5 6 6 7 7 7 8 8 9 9

Return iterator from std::upper_bound

#include <iostream>
using std::cout;
using std::endl;
#include <algorithm>
#include <vector>
#include <iterator>
int main()
{
   int a1[ 10 ] = { 2, 2, 4, 4, 4, 6, 6, 6, 6, 8 };
   std::vector< int > v( a1, a1 + 10 );
   std::ostream_iterator< int > output( cout, " " );
   std::copy( v.begin(), v.end(), output );
   // determine upper-bound insertion point for 6 in v
   std::vector< int >::iterator upper;
   upper = std::upper_bound( v.begin(), v.end(), 6 );
   cout << "::"<< ( upper - v.begin() );
   return 0;
}
2 2 4 4 4 6 6 6 6 8 ::9

Upper bound of 6

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
   const int SIZE = 10;
   int a1[] = { 2, 2, 4, 4, 4, 6, 6, 6, 6, 8 };
   vector< int > v( a1, a1 + SIZE );
   vector< int >::iterator upper;
   upper = upper_bound( v.begin(), v.end(), 6 );
   cout << "\nUpper bound of 6 is element " << ( upper - v.begin() ) << " of vector v";
   return 0;
}

Use upper_bound to locate the last point

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
   const int SIZE = 10;
   int a1[] = { 2, 2, 4, 4, 4, 6, 6, 6, 6, 8 };
   vector< int > v( a1, a1 + SIZE );
   vector< int >::iterator upper;
   cout << "\n\nUse upper_bound to locate the last point\n"
        << "at which 7 can be inserted in order";
   upper = upper_bound( v.begin(), v.end(), 7 );
   cout << "\n   Upper bound of 7 is element " 
        << ( upper - v.begin() ) << " of vector v";
   return 0;
}