就是这个，一英里。

bool once(uintptr_t val) {
    return visited.emplace(val).second;
}

怎么会不是这样呢?

https://godbolt.org/z/9zP77jqMc

func5(unsigned long):
        sub     rsp, 24
        mov     QWORD PTR [rsp+8], rdi
        lea     rsi, [rsp+8]
        mov     edi, OFFSET FLAT:visited2
        call    std::pair<std::_Rb_tree_iterator<unsigned long>, bool> std::_Rb_tree<unsigned long, unsigned long, std::_Identity<unsigned long>, std::less<unsigned long>, std::allocator<unsigned long> >::_M_emplace_unique<unsigned long&>(unsigned long&)
        add     rsp, 24
        mov     eax, edx
        ret

Just to clarify, the reason why there is no member like contains() in these container types is because it would open you up to writing inefficient code. Such a method would probably just do a this->find(key) != this->end() internally, but consider what you do when the key is indeed present; in most cases you'll then want to get the element and do something with it. This means you'd have to do a second find(), which is inefficient. It's better to use find directly, so you can cache your result, like so:

auto it = myContainer.find(key);
if (it != myContainer.end())
{
    // Do something with it, no more lookup needed.
}
else
{
    // Key was not present.
}

当然，如果你不关心效率，你总是可以自己滚动，但在这种情况下，你可能不应该使用c++…；）

另一种简单地判断元素是否存在的方法是检查count()

if (myset.count(x)) {
   // x is in the set, count is 1
} else {
   // count zero, i.e. x not in the set
}

然而，大多数时候，我发现自己需要访问元素，无论我在哪里检查它的存在。

所以我还是要找到迭代器。当然，最好也把它和end进行比较。

set< X >::iterator it = myset.find(x);
if (it != myset.end()) {
   // do something with *it
}

C + + 20

在c++ 20中，set获得一个contains函数，因此如下所述成为可能:https://stackoverflow.com/a/54197839/895245

if (myset.contains(x)) {
  // x is in the set
} else {
  // no x 
}

我能够为std::list和std::vector编写一个通用的包含函数，

template<typename T>
bool contains( const list<T>& container, const T& elt )
{
  return find( container.begin(), container.end(), elt ) != container.end() ;
}

template<typename T>
bool contains( const vector<T>& container, const T& elt )
{
  return find( container.begin(), container.end(), elt ) != container.end() ;
}

// use:
if( contains( yourList, itemInList ) ) // then do something

这样可以稍微清理一下语法。

但是我不能使用模板模板参数魔术使此工作任意stl容器。

// NOT WORKING:
template<template<class> class STLContainer, class T>
bool contains( STLContainer<T> container, T elt )
{
  return find( container.begin(), container.end(), elt ) != container.end() ;
}

任何关于改进上一个答案的评论都是很好的。

我使用

if(!my_set.count(that_element)) //Element is present...
;

但它的效率不如

if(my_set.find(that_element)!=my_set.end()) ....;

我的版本只是节省了我写代码的时间。对于竞争性编码，我更喜欢这种方式。

写你自己的:

template<class T>
bool checkElementIsInSet(const T& elem, const std::set<T>& container)
{
  return container.find(elem) != container.end();
}