我好奇，如果有一些类似于cava中的Java哈希表。即，一个具有快速查看的数据结构，因为我将只对其运行.contains（e）。同样，如果你可以启发我如何做一个.contains（）在你提出的任何数据结构，我会非常感谢。 O，请不要张贴只看看c ++ docs，因为我已经这样做，并发现他们负担。
解决方案您可以使用 std :: unordered_set<> §23.5.6），其 找到 方法（执行查找）作为O（1）的平均复杂度：
#include< iostream> #include< unordered_set> int main（） { std :: unordered_set< int> example = {1，2，3，4}; auto search = example.find（2）; if（search！= example.end（））{ std :: cout< Found< （* search）<< '\\\'; } else { std :: cout< 未找到\\\; } }EDIT： p>
根据@Drew Dormann的建议，您也可以使用 count ，其平均复杂度为O（1）：
#includeI as curious if there was something akin the the Java hashset in c++. I.e a data structure with fast look, as I will only be running .contains(e) on it. Likewise, if you could enlighten me on how to do a .contains() on whatever data structure you propose, I would be very appreciative. O, please do not post just look at the c++ docs as I have already done so and find them burdensome.
解决方案You can use std::unordered_set<> (standard § 23.5.6), its find method (to do a lookup) as an average complexity of O(1) :
#include <iostream> #include <unordered_set> int main() { std::unordered_set<int> example = {1, 2, 3, 4}; auto search = example.find(2); if(search != example.end()) { std::cout << "Found " << (*search) << '\n'; } else { std::cout << "Not found\n"; } }EDIT:
As suggested by @Drew Dormann, you can alternatively use count, which also has a average complexity of O(1):
#include <iostream> #include <unordered_set> int main() { std::unordered_set<int> example = {1, 2, 3, 4}; if(example.count(2)) { std::cout << "Found\n"; } else { std::cout << "Not found\n"; } }
更多推荐
Java HashSet等价于c ++
发布评论