Reputation: 11
I have a std::vector<int> numList
and it contains {9,6,3,3,2,1}
. What I want, is during the loop, once we hit 9 % 3 == 0
, I want to put 9
into a new vector v1
and 3
into v2
, while erasing these numbers from the original list. And repeat the process every time the mod results in 0
.
I have this but it crashes:
for(i = 0; i < listSize; i++)
{
for(j = 0; j < listSize; j++)
{
if(i != j)
{
remainder = numList[i] % numList[j];
if(numList[i] % numList[j] == 0)
{
//cout<< numList[i] << " " << numList[j]<<endl;
v1.push_back(numList[i]);
v2.push_back(numList[j]);
numList.erase(numList.begin() + i);
numList.erase(numList.begin() + j);
}
}
}
}
Upvotes: 0
Views: 140
Reputation: 19607
You'll have to discard listSize
and use numList.size()
, that will give you the current size. You might as well delete the whole remainder = numList[i] % numList[j];
, while we're at it. I guess you're not using remainder
afterwards, remove it completely.
Important:
i
and j
in the cycle where erasing occurred - you don't want to skip any elements.1
in among the elements that will pair up with anything. Fix the condition.To conclude, this is not cool:
int i; // just absurd
// list of variable declarations that are not needed right now, or not needed at all
for(i = 0; i < listSize; i++)
This is much better:
for(int i = 0; i < numList.size(); i++)
Upvotes: 1
Reputation: 310980
I think you need something like the following
#include <iostream>
#include <iterator>
#include <vector>
int main()
{
std::vector<int> v = { 9, 6, 3, 3, 2, 1 };
std::vector<int> v1;
std::vector<int> v2;
for ( std::vector<int>::size_type i = 0; i < v.size(); )
{
std::vector<int>::size_type j = i + 1;
while ( j < v.size() && v[i] % v[j] ) j++;
if ( j != v.size() )
{
v1.push_back( v[i] );
v2.push_back( v[j] );
v.erase( std::next( v.begin(), j ) );
v.erase( std::next( v.begin(), i ) );
}
else
{
i++;
}
}
for ( int x : v1 ) std::cout << x << ' ';
std::cout << std::endl;
for ( int x : v2 ) std::cout << x << ' ';
std::cout << std::endl;
}
The program output is
9 6 2
3 3 1
Take into account that the index in the inner loop can not be less than the index in the outer loop for matching elements.
Upvotes: 0