c++ - 插入器迭代器，可根据谓词插入不同的容器

Question

可以说我有一个数字 vector ，我想创建两个 vector 以将奇数和偶数分开。轻松实现以下目的:

std::vector<int> odds;
std::vector<int> evens;
std::vector<int> numbers;
for (int number : numbers) {
    if isOdd(number)
        odds.push_back(number);
    else
        evens.push_back(number);
}

我想知道是否有任何类型的inserter_iterator可以做到这一点，所以我可以写一些

std::vector<int> odds;
std::vector<int> evens;
std::vector<int> numbers;

auto pred = [](int i) { return isOdd(i) ? True : False;};
auto identity = [](int i) {return i;};
std::transform(std::begin(numbers), std::end(numbers), some_inserter(odd, evens, pred), identity);

这只是出于好奇，我试图了解算法和迭代器的工作方式。基于范围的解决方案也是有效的。

Answer 1

您可以使用std::partition_copy算法，如下所示https://en.cppreference.com/w/cpp/algorithm/partition_copy

int main() {
    std::vector vec = {1, 2, 3, 4, 5, 6};//the vector to be partitioned
    std::vector<int> odds(vec.size());//will hold odds (to have the sam size as the original (the maximum case) )
    std::vector<int> evens(vec.size());//will hold evens

    auto [oddsEnd, evensEnd] =
            std::partition_copy(vec.begin(), vec.end(), odds.begin(), evens.begin(), [](int i){return i%2!=0;});//will copy at the front (won't insert at the back)
    odds.erase(oddsEnd, odds.end());//to erase the undesired excess
    evens.erase(evensEnd, evens.end());//to erase the undesired excess

}

或使用 std::back_inserter https://en.cppreference.com/w/cpp/iterator/back_inserter(请参阅评论)

int main() {
    std::vector vec = {1, 2, 3, 4, 5, 6};//the vector to be partitioned
    std::vector<int> odds;//will hold odds
    std::vector<int> evens;//will hold evens

    std::partition_copy(vec.begin(), vec.end(), std::back_inserter(odds), std::back_inserter(evens), [](int i){return i%2!=0;});//will insert

}