Question

让x和y是两个排序的向量。我想找出三个中的哪个是正确的

A. All elements of `y` are present in `x`
B. Some but not all elements of `y` are present in `x`
C. No element of `y` is present in `x`
U. Undefined (because `y` is empty)

实现这一目标的一种简单方法是

template<typename T>
char f(std::vector<T> x, std::vector<T> y)
{
    if (y.size() == 0)
    {
        return 'U';
    }

    bool whereAnyElementFound = false;
    bool whereAnyElementNotFound = false;
    for (T& y_element : y)
    {
        bool isElementFound = std::find(x.begin(),x.end(),y_element) != x.end();
        if (isElementFound)
        {
            whereAnyElementFound = true;
            if (whereAnyElementNotFound)
            {
                return 'B';
            }
        } else
        {
            whereAnyElementNotFound = true;
            if (whereAnyElementFound)
            {
                return 'B';
            }
        }
    }
    if (whereAnyElementFound)
    {
        return 'A';
    } else if (whereAnyElementNotFound)
    {
        return 'C';
    }
    abort();
}

该功能将以下输入正确匹配到输出

inputs: x = {1,2,4,5} y = {2,5}
output: A

inputs: x = {1,2,4,5} y = {2,7}
output: B

inputs: x = {1,2,4,5} y = {6,7}
output: C

inputs: x = {1,2,4,5} y = {}
output: U

但是，该方法没有利用两个向量都被排序的事实。对于较大的向量，如何使此功能更快？

Answer 1

您可以使用std::set_intersection，以支付O(N)的额外空间。它具有O(2(N1+N2-1))的复杂性，并生成两个向量之间所有公共元素的“集合”。然后，您可以检查新的“集合”的大小以找出A，B，C和U。

int main()
{
    std::vector<int> v1{1,2,3,4,5,6,7,8};
    std::vector<int> v2{5,7,9,10};

    std::vector<int> intersection;

    std::set_intersection(v1.begin(), v1.end(),
                          v2.begin(), v2.end(),
                          std::back_inserter(intersection));
    if (intersection.size() == v2.size() && v2.size() > 0)
        std::cout << "A";
    else if (intersection.size() > 0)
        std::cout << "B";
    else if (intersection.size() == 0 && v2.size() > 0)
        std::cout << "C";
    else
        std::cout << "U";
}

是否所有x元素都存在于y中（排序的向量）？

1 个答案: