I have a vector a storing values [0 1 2 3 5] and other vector removelist storing the indexes to be removed [0 1 2] in order to leave [3 5] at the end. When I'm implementing the following code, it would remove items unexpectedly since the vector a will be changing order during the process. Is there any way for me to achieve my target?
for (int i = 0; i<removelist.size() ; i++) a.erase(a.begin() + removelist[i]); Reverse the order you remove values, i.e. use the reverse iterators of removelist. This of course relies on removelist being sorted.
Perhaps something like
std::sort(removelist.begin(), removelist.end()); // Make sure the container is sorted for (auto &i = removelist.rbegin(); i != removelist.rend(); ++ i) { a.erase(a.begin() + *i); } 
Not necessarily more efficient, but you can do this without sorting using remove_if:
auto& rm = removelist; // for brevity a.erase(remove_if(begin(a), end(a), [&](int i) { auto idx = distance(begin(v), find(begin(v), end(v), i)); return find(begin(rm), end(rm), idx) != end(rm); }, end(a)); 
find(begin(v), end(v), i) will lead to unexpected result when there are the elements in the vector v are not unique.The solution to this is to copy the elements you want to keep to a new vector:
// pseudocode: vector tmp; tmp.reserve(a.size() - removelist.size()); for (i=0; i<a.size(); ++i) { if (i not in removelist) { tmp.push_back(a[i]); } } a.swap(tmp); Notes:
a also avoid the index shift in your approach.removelst are sorted, this can be implemented a bit more efficiently.
Adapted from @Yam Marcovic's answer, not using find but use the exact address to find the index:
a.erase(std::remove_if(a.begin(), a.end(), [&](const int& i) { auto idx = ((void*)&i - (void*)&*a.begin()); return std::find(removelist.begin(), removelist.end(), idx) != removelist.end(); }), a.end()); 
removelistordered?avector is ordered as he removes it by indices not by values. But his example is buggy (there is no index 5 in a vector with size 5)