简单来说,unordered_key非常的高效,使用值本身作为键值 Example #include<iostream>#include<unordered_set>using namespace std;intmain(){// 1. initialize a hash setunordered_set<int>hashset;// 2. insert a new keyhashset.insert(1);hashset.insert(2);hashset.insert(3);// 3. delete a keyhash...
insert(s[i]); } if(set.count(ch) == 1){ cout<<"char "<<ch<<" is present so it will print: "; return true; } cout<<"char "<<ch<<" is not present so it will print: "; return false; } int main() { cout << isPresent("tutorialspoint", 't') << '\n'; return 0;...
C++11 unordered_set::insert C++11 unordered_set::key_eq C++11 unordered_set::load_factor C++11 unordered_set::max_bucket_count C++11 unordered_set::max_load_factor C++11 unordered_set::max_size C++11 unordered_set::operator= C++11 unordered_set::rehash C++11 unordered_set::re...
insert,emplace,emplace_hintOnly if causes rehash eraseOnly to the element erased Notes The swap functions do not invalidate any of the iterators inside the container, but they do invalidate the iterator marking the end of the swap region. ...
t.insert(b.begin(), b.end());returnt; }intmain(intargc,char**argv){ unordered_set<string> first1;unordered_set<string>first2( {"one","two","three"} );unordered_set<string>first3( {"red","green","blue"} );unordered_set<string>first4( first2 );unordered_set<string>first5( c...
Complexity Constant. Iterator validity All iterators, pointers and references remain valid, but now are referring to elements in the other container, and iterate in it. non-member overloads: operators (unordered_set) swap (unordered_set)
unordered_set::insert unordered_set::insert_range (C++23) unordered_set::emplace unordered_set::emplace_hint Lookup unordered_set::count unordered_set::find unordered_set::contains (C++20) unordered_set::equal_range Bucket interface unordered_set::begin(size_type)unordered_set::cbegin(size_type...
() ; return result ; } void insert( const std::vector<std::size_t>& srce, std::vector<std::size_t>& dest ) { dest = srce ; std::sort( std::begin(dest), std::end(dest) ) ; } void insert( const std::vector<std::size_t>& srce, std::unordered_set<std::size_t>& ...
But in the worst-case scenario, the unordered_map is slower than a map because the worst time complexity of all the operations in an unordered_map (O(n)) is greater than the time complexity for all the operations in a map (O(log n)). ...
Describe the bug If the underlying strings that an unordered_set<string_view> contains are invalidated, a call to unordered_set::clear() will succeed, but the second insert() after that will cause a read access violation when the fast pa...