Good way and bad way to remove elements in a list
/* 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 <list>
#include <algorithm>
using namespace std;
int main()
{
list<int> coll;
// insert elements from 6 to 1 and 1 to 6
for (int i=1; i<=6; ++i) {
coll.push_front(i);
coll.push_back(i);
}
// remove all elements with value 3
// - poor performance
coll.erase (remove(coll.begin(),coll.end(),
3),
coll.end());
// remove all elements with value 4
// - good performance
coll.remove (4);
}Remove all elements with the same value
#include <iostream>
using std::cout;
using std::endl;
#include <list> // list class-template definition
#include <algorithm> // copy algorithm
#include <iterator> // ostream_iterator
int main()
{
int array[ 4 ] = { 2, 6, 4, 8 };
std::list< int > values; // create list of ints
std::list< int > otherValues; // create list of ints
std::ostream_iterator< int > output( cout, " " );
// insert items in values
values.push_front( 1 );
values.push_front( 3 );
values.push_back( 4 );
values.push_back( 1 );
values.push_back( 2 );
values.push_back( 3 );
cout << "values contains: ";
std::copy( values.begin(), values.end(), output );
values.remove( 1 ); // remove all 1s
cout << "\n\nvalues contains: ";
std::copy( values.begin(), values.end(), output );
cout << endl;
return 0;
}values contains: 3 1 4 1 2 3
values contains: 3 4 2 3
Remove element from a list
#include <iostream>
#include <list>
using namespace std;
void show(const char *msg, list<char> lst);
int main() {
list<char> lstA;
lstA.push_back("A");
lstA.push_back("F");
lstA.push_back("B");
lstA.push_back("R");
// Remove A and H.
lstA.remove("A");
lstA.remove("H");
show("lstA after removing A and H: ", lstA);
cout << endl;
return 0;
}
void show(const char *msg, list<char> lst) {
list<char>::iterator itr;
cout << msg << endl;
for(itr = lst.begin(); itr != lst.end(); ++itr)
cout << *itr << endl;
}