I made it work for me by recasting my priority_queue pointer to a pointer to a vector (as my priority_queue uses vector as its container). I had the same problem, where I wanted to iterate over a priority queue without dequeuing (hence destroying my queue). The priority-queue container has the following declaration: template< typename ValueType, typename CmpFn std::less, typename Tag.Finally, it runs in O(n) time instead of the commonly seen O(n log n). Also it is compact, trivial to implement, does not change the original object and does not require additional memory. This approach has the advantage of not changing the data structure used : if a binary heap is suited for your application, there is no point in using a multiset. If you want to iterate over the priority queue to search for whether it contains a value then maybe create a wrapper priority queue and use a hash set to keep track of what you have in the queue. Std::cout << std::endl // outputs "42 16 5 1 3" begin () pointer (like vector would do) that you can use to iterate over it. Q.emplace(1) q.emplace(16) q.emplace(5) q.emplace(42) q.emplace(3) Ĭonst T* base = &q.top() // root of the binary heap If these conditions are met, the following code can be used : #include Parameters (none) Return value Reference to the top element as if obtained by a call to c.front() Complexity Constant. If default comparison function is used, the returned element is also the greatest among the elements in the queue. This element will be removed on a call to pop (). >classpriorityqueue A priority queueis a container adaptor that provides constant time lookup of the largest (by default) element, at the expense of logarithmic insertion and extraction. Returns reference to the top element in the priority queue. ⢠it is not necessary to go through the elements in sorted order. classT, classContainer std::vector, classCompare std::less.the underlying container is a vector, and.The operation can be performed in linear time with respect to the number of elements, provided that : Some even advise to make a copy of the structure. Most existing answers either suggest using a different data structure or performing a sequence of O(log n) operations ( n being the size of the priority queue).
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |