C++ 在 std::map 中搜索特定值
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Search for specific value in std::map
提问by B?ови?
Possible Duplicates:
Checking value exist in a std::map - C++
How to traverse a stl map/vector/list/etc?
Hello,
你好，
Is it possible to search for specific value in std::map, not knowing the key? I know I could iterate over whole map, and compare values, but it is possible to do using a function from std algorithms?
是否可以在不知道键的情况下在 std::map 中搜索特定值？我知道我可以遍历整个地图并比较值，但是可以使用 std 算法中的函数吗？
采纳答案by Platinum Azure
Will this help? STL find_if
这会有所帮助吗？STL find_if
You need to have some sort of predicate, either a function pointer or an object with operator()implemented. Said predicate should take just one parameter.
您需要有某种谓词，函数指针或已operator()实现的对象。所述谓词应仅采用一个参数。
回答by icecrime
Well, you could use std::find_if:
好吧，你可以使用std::find_if：
int main()
{
typedef std::map<int, std::string> my_map;
my_map m;
m.insert(std::make_pair(0, "zero"));
m.insert(std::make_pair(1, "one"));
m.insert(std::make_pair(2, "two"));
const std::string s("one");
const my_map::const_iterator it = std::find_if(
m.begin(), m.end(), boost::bind(&my_map::value_type::second, _1) == s
);
}
But that's just slightly better than a hand-crafted loop : it's still O(n).
但这只是比手工制作的循环稍微好一点：它仍然是O(n).
回答by Steve Townsend
You could use Boost.Bimapif you want to index on values as well as keys. Without this or similar, this will have to be done by brute force (=> scan the mapby hand).
如果您想对值和键进行索引，则可以使用Boost.Bimap。如果没有这个或类似的，这将不得不通过蛮力来完成（=>map手动扫描）。
Boost.Bimap is a bidirectional maps library for C++. With Boost.Bimap you can create associative containers in which both types can be used as key.
Boost.Bimap 是 C++ 的双向映射库。使用 Boost.Bimap 您可以创建关联容器，其中两种类型都可以用作键。
回答by Marcelo Cantos
There are (awkward) ways to do this using standard functions (e.g., std::find_if), but these still involve iterating over the whole map. Boost.Bimapwill provide efficient indexing in both directions, and you can go even further with Boost.Multi-Index.
有（笨拙的）方法可以使用标准函数（例如 ）来做到这一点std::find_if，但这些方法仍然涉及对整个地图进行迭代。Boost.Bimap将在两个方向上提供高效的索引，您可以使用Boost.Multi-Index走得更远。
