std :: vector在最后移动反向迭代器

Question

I have this situation:

for(auto it = vec.rbegin(); it != vec.rend(); ++it)
{
    if(condition(it))
    {
        //Move it at end;
        break;
    }
}

What is the most efficient/elegant way to move *it at the end of vec ?

EDIT: *it , not it

EDIT1: Now I use:

auto val = *it; 
vec.erase((++it).base());
vec.push_back(val);

But I don't think is very efficient...

Answer 1

To move it to the end:

for(auto it = vec.rbegin(); it != vec.rend(); ++it)
{
    if(condition(it))
    {
        auto val = *it;
        vec.erase((++it).base());
        vec.push_back(val);
        break;
    }
}

To swap with the last element:

for(auto it = vec.rbegin(); it != vec.rend(); ++it)
{
    if(condition(it))
    {
        auto it2 = --vec.end();
        auto val = *it;
        *it = *it2;
        *it2 = val;
        break;
    }
}

Answer 2

If prefer standard algorithms over manual loops. So I'd use std::find_if .

If you need to preserve the order of the elements, moving to the end can be done using std::rotate :

auto rit = std::find_if(vec.rbegin(), vec.rend(), [](int i){ return i == 6; });
if (rit != vec.rend()) {
    auto it = rit.base();
    std::rotate(it-1, it, vec.end());
}

If you don't need to keep the order, you can use std::iter_swap :

auto it = std::find_if(vec.rbegin(), vec.rend(), [](int i){ return i == 6; });
if (it != vec.rend())
    std::iter_swap(it, vec.rbegin());

Demo