Re: removing from set - does it have to be so ugly?

From:
peter koch larsen <peter.koch.larsen@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Tue, 24 Nov 2009 10:03:39 CST
Message-ID:
<f7febfc7-76e8-49ca-ac3f-442c452016e0@m20g2000vbp.googlegroups.com>
On 24 Nov., 06:41, n00m <n...@narod.ru> wrote:

On Nov 23, 11:42 pm, Ulrich Eckhardt <eckha...@satorlaser.com> wrote:

The correct code
requires careful use of a postfix increment operator:

      set<int>::iterator rit = r.begin();
      while (rit != r.end()) {
          if (*rit % 2 == 1)
              r.erase(rit++);
          else
              ++rit;
      }


Seems you are goddam right
but ouch I can't still tell the difference
and personally I'd better use the commented method:

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <set>
using namespace std;

void print(int elem) {cout << elem << endl;}

bool is_odd(int x) {return x % 2 != 0;}

int main() {
     set<int> r;
     int tmp[] = {0,1,-5,-3,8,3,2,4,5,6,7,8,9,11,22,44,55};
     r.insert(tmp, tmp + sizeof(tmp)/sizeof(int));

     set<int>::iterator rit = r.begin();
     while (rit != r.end()) {
         if (is_odd(*rit))
             r.erase(rit++);
         else
             ++rit;
     }

     /*
     while (true) {
         set<int>::iterator rit = find_if(r.begin(), r.end(), is_odd);
         if (rit == r.end()) break;
         r.erase(rit);
     }
     */

     for_each(r.begin(), r.end(), print);

system("pause");
return 0;

}


The problem with your solution is that you traverse the set once for
each element deleted and then one more time. If the number of elements
to delete is proportional to the total number of elements, your
solution becomes O(n*n) instead of O(n).

/Peter

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