如何以无序顺序（C ++）测试一项成员资格？

Question

I've used python for a long time, and I'm just beginning to use C++

In python, if one has a set or a dictionary it is relatively easy to get a boolean value indicating whether or not a particular item is in that sequence using the keyword. ie

a = set(2,4,3)   
if 4 in a  
print "yes, 4 is in a, thank you for asking!"

it's much more efficient than doing this:

a = [2,3,4]  
for number in a
>if number == 4  
>>return "yes, 4 is in a, thank you for asking!"

is there a way to do make a membership test simple and efficient in cpp or do you always have to iterate through some ordered sequence?

Answer 1

You have functionality like this in std::set and tr1::unordered_set (not yet in C++ standard).

#include <set>
#include <cstdio>

int main() {
     std::set<int> s;
     s.insert(1);        
     s.insert(2);
     s.insert(4);
     if (s.find(4) != s.end())
          puts("4 found!");
     return 0;
}

In reality, if your data set is small, linear search may still be the faster option.

Answer 2

Get to know the C++ Standard Template Library. The set class (and others) has a find() method that will return an iterator to an item in the set if it exists.

Answer 3

看一下STL提供的容器及其性能特征。

Answer 4

Python's method is not "much more efficient" because you don't know the complexity of in construct.

In C++ there are many methods of storing data. Binary trees are best for searching.

If you work with numbers like 2,3,4 etc, you may consider having an array of bools, and simply see if array[4] == true

Answer 5

template<typename T>
bool contains(const std::set<T>& a, const T& value) {
    return a.find(value) != a.end();
}

if (contains(a, 4)) {
    std::cout << "A contains 4\n";
}

Answer 6

std::find can determine whether or not an element exists in any unordered container or sequence. If the sequence is unordered and not stored in a specialized container designed for lookups, it's unlikely you're going to do any better than O(N) .