Java HashSet等价于c ++ [英] Java HashSet equiv in c++

问题描述

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）：

  #include  #include< unordered_set> 
 
 int main（）
 {
 std :: unordered_set< int> example = {1，2，3，4}; 
 
 if（example.count（2））{
 std :: cout< Found\\\
; 
} 
 else {
 std :: cout< 未找到\\\
; 
} 
} 
  

I 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";
    }
}

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