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Pen72 (74)
The way to swing the list follows the algorithm below :

1Pick the smallest integer from s and append it to the result.
2Pick the smallest integer from which is greater than the last appended integer to the result and append it.
3Repeat step 2 until you cannot pick more integers.
4Pick the largest integer from s and append it to the result.
5Pick the largest integer from s which is smaller than the last appended integer to the result and append it.
6Repeat step 5 until you cannot pick more integer.
7Repeat the steps from 1 to 6 until you pick all integers from s .


Input Format

The first line has a integer meaning there are n lists.

The next n line, each line has a list. The first integer l of a line is the length of the list. And followed integers are the elements of the list.

I've tried modifying the iterators, while the code below shows no output.

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#include <cmath>
#include <cstdio>
#include <vector>
#include <queue>
#include <list>
#include <iostream>
#include <algorithm>
#include <set>
#include <iterator>
using namespace std;


int main() 
{
   int n,i;
    cin>>n;
    while(n--)
    {
       int L,val;
        cin>>L;  
        vector<int> A(L);
        vector<int>::iterator it,itr;
        A.clear();

       for(auto& a:A)
           cin>>a;
  
        sort(A.begin(),A.end());
 
       while(!A.empty())
       {
          val=*(A.begin());
          cout<<val<<" ";
          
          it=A.begin();
          for( i=0;i<A.size();i++)
          {
            
            if(*(it+i)>val)
            {
                val=*(it+i);
                cout<<val<<" ";
                itr=it+i;
                A.erase(itr);
                it--;
                
            }
          }   
           
          val=*(A.end()-1);
          cout<<val<<" ";
        
         it=(A.end()-1);
         for(i=0;i<A.size(); i++)
        {
          
            if(*(it-i)<val)
            {
                val=*(it-i);
                cout<<val<<" ";
                itr=it-i;
                A.erase(itr);
                it--;
                
            }
        }

       }
        cout<<"\n";
        
    }
    return 0;
}
Last edited on
seeplus (6591)
seeplus (6591)
L35 it is set to .begin()
L45 - --it it is now invalid as it points before .begin() !

Also don't forget that for std::vector .erase() invalidates any iterator pointing to or following the value(s) removed. If iterators are used, then .erase() returns a valid iterator. See http://www.cplusplus.com/reference/vector/vector/erase/

Again - use the debugger to trace through the code and identify issue(s).
Last edited on
Pen72 (74)
But if I didn't do it-- in line 45 ,when I do A.erase(itr) in line 44 wouldn't the total number of iterators -1? that's why I thought there should be a it-- since it=A.begin() in line 35 was set before them.
coder777 (8444)
Using the iterator on line 45 is illegal due to the erase on line 44. Why do you use the iterator at all? See:
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          for( i=0;i<A.size();i++)
          {
            
            if(A[i]>val)
            {
                val=A[i];
                cout<<val<<" ";
                A.erase(A.begin() + i);
                --i; // Not skipping an element
            }
          }   
          if(not A.empty())
            val=A.back();
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