What is time complexity for find method in a set in c++?
set<int> s;
s.insert(1);
s.insert(2);
...
s.insert(n);
I wonder how much time it take开发者_Python百科s for s.find(k)
where k
is a number from 1..n?
I assume it is log(n). Is it correct?
O( log N ) to search for an individual element.
§23.1.2 Table 69
expression return note complexity
a.find(k) iterator; returns an iterator pointing to an logarithmic
const_iterator element with the key equivalent to k,
for constant a or a.end() if such an element is not
found
The complexity of std::set::find()
being O(log(n))
simply means that there will be of the order of log(n)
comparisons of objects stored in the set
.
If the complexity of the comparison of 2 elements in the set is O(k)
, then the actual complexity, would be O(log(n)∗k)
.
this can happen for example in case of set of strings (k would be the length of the longest string) as the comparison of 2 strings may imply comparing most of (or all) of their characters (if they start with the same prefix or are equal)
The documentation says the same:
Complexity
Logarithmic in size.
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