c++ - vector.erase 和 std::remove 自定义 vector

Question

我有一个整数对 vector 。假设 Itr 作为这个 vector 的迭代器。我想遍历 vector 并决定是否从 vector 中删除元素。如果 vector 的元素是 9001,3 那么我想从 itr->first 是 9001 的 vector 中删除所有元素(不管是什么 itr ->第二个是)。

问题:

1. 如何删除这个 int 对 vector 。以下是行不通的:vec.erase(std::remove(vec.begin(), vec.end(), Itr->first=9001), vec.end());
2. 不是给 vec.begin() 到 vec.end() 的范围，我可以给vec.begin as (当前元素是 vector 指针)- 10 和ve.end 为 ( vector 指向的当前元素)+10 ？

示例:vec.erase(std::remove(Itr-10, Itr+10, Itr->first=9001), vec.end());

如果 vector 大小本身小于 10，Itr+10 或 Itr-10 可能会导致段错误。那么如何处理这种情况？

    vector<pair<int,int> > vec;

    vec.push_back(pair<int,int>(9001,1));
    vec.push_back(pair<int,int>(9001,2));
    vec.push_back(pair<int,int>(9001,3));
    vec.push_back(pair<int,int>(9001,4));
    vec.push_back(pair<int,int>(9002,1));
    vec.push_back(pair<int,int>(9002,2));
    vec.push_back(pair<int,int>(9002,3));
    vec.push_back(pair<int,int>(9002,4));
    vec.push_back(pair<int,int>(9002,5));


    vector<pair<int,int> >::iterator Itr;
    for(Itr=vec.begin();Itr!=vec.end();++Itr)
            cout<<vecItr->first<<" "<<vecItr->second;

//  vec.erase(std::remove(Itr-10, Itr+10, Itr->first=9001), vec.end()); //This doest work

    for(Itr=vec.begin();Itr!=vec.end();++Itr)
            cout<<Itr->first<<" "<<Itr->second;

Answer 1

How do i remove this vector of int pairs. The following wouldn't work:

将 lambda 或自定义比较器与 remove_if 算法一起使用:

vec.erase(std::remove_if(vec.begin(), vec.end(), 
                         [](auto& elem){ return elem.first == 9001;} ),
          vec.end());

使用自定义比较器:

struct elem_equals
{
    typedef std::pair<int,int> elem_t
    const int value;

    elem_equals(int v) : value(v) {}

    bool operator()(elem_t& elem) 
    {
        return elem.first == value;
    }
};

//...

vec.erase(std::remove_if(vec.begin(), vec.end(), 
                         elem_equals(9001) ),
          vec.end());

Instead of giving a range as vec.begin() to vec.end() is it possible for me to give vec.begin as (current element being pointer by vector) - 10 and ve.end as (current element being pointed by vector)+10 ?

是的。 vector 迭代器支持指针运算，所以很容易。

It might quiet be possible that Itr+10 ot Itr-10 will result in a segmentation fault if the vector size itself is less than 10. So how to deal with this situation?

如果迭代器之前或之后没有足够的元素，则限制你的范围:

//iter is an iterator to vector
//vec is instance of std::vector
//replace auto with std::vector<std::pair<int,int> >::iterator for non C++11 compilers
auto begin = iter - std::min(10, iter - vec.begin());
auto end = iter + std::min(10, vec.end() - iter);