I was curious if there was something akin the Java HashSet in C++? I.e. a data structure with a 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 found them burdensome.
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Milan
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user1352683
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2`std::unordered_set` would be the closest โ quantdev Jul 09 '14 at 01:47
1 Answers
20
You can use std::unordered_set<>
(standard ยง 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";
}
}

quantdev
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4I'd recommend `count()` for emulating contains(). That 2 line if becomes simply `if(example.count(2))` โ Drew Dormann Jul 09 '14 at 01:57
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4The function was named `count()` to be consistent with multisets, but for `unordered_set` it will only return `1 (true)` or `0 (false)` โ Drew Dormann Jul 09 '14 at 02:09