如何仅从forward_list中删除单个元素,如何有效地将其删除?
好吧,我认为这个问题几乎可以总结出来。我有一个独特商品的forward_list,想从中删除一个商品:
std::forward_list<T> mylist;// fill with stuff
mylist.remove_if([](T const& value)
{
return value == condition;
});
我的意思是,这种方法效果很好,但是效率不高,因为一旦找到并删除了该项目,它就会继续搜索。有更好的方法还是我需要手动进行?
回答:
如果你只是想删除的第一场比赛,你可以使用std::adjacent_find
其次成员erase_after
#include <algorithm>#include <cassert>
#include <forward_list>
#include <iostream>
#include <ios>
#include <iterator>
// returns an iterator before first element equal to value, or last if no such element is present
// pre-condition: before_first is incrementable and not equal to last
template<class FwdIt, class T>
FwdIt find_before(FwdIt before_first, FwdIt last, T const& value)
{
assert(before_first != last);
auto first = std::next(before_first);
if (first == last) return last;
if (*first == value) return before_first;
return std::adjacent_find(first, last, [&](auto const&, auto const& R) {
return R == value;
});
}
int main()
{
auto e = std::forward_list<int>{};
std::cout << std::boolalpha << (++e.before_begin() == end(e)) << "\n";
std::cout << (find_before(e.before_begin(), end(e), 0) == end(e)) << "\n";
auto s = std::forward_list<int>{ 0 };
std::cout << (find_before(s.before_begin(), end(s), 0) == s.before_begin()) << "\n";
auto d = std::forward_list<int>{ 0, 1 };
std::cout << (find_before(d.before_begin(), end(d), 0) == d.before_begin()) << "\n";
std::cout << (find_before(d.before_begin(), end(d), 1) == begin(d)) << "\n";
std::cout << (find_before(d.before_begin(), end(d), 2) == end(d)) << "\n";
// erase after
auto m = std::forward_list<int>{ 1, 2, 3, 4, 1, 3, 5 };
auto it = find_before(m.before_begin(), end(m), 3);
if (it != end(m))
m.erase_after(it);
std::copy(begin(m), end(m), std::ostream_iterator<int>(std::cout, ","));
}
找到匹配项后,该操作将立即停止。请注意,该adjacent_find
接受一个二进制谓词,并且通过仅比较第二个参数,我们在要删除的元素之前获得了一个迭代器,因此erase_after
实际上可以将其删除。复杂性就是O(N)
这样,您将无法获得比它更有效的效果。
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