deque

What is a data structure that has O(1) for append, prepend, and retrieve element at any location?

China☆狼群 提交于 2019-12-10 02:38:20
问题 I'm looking for Java solution but any general answer is also OK. Vector/ArrayList is O(1) for append and retrieve, but O(n) for prepend. LinkedList (in Java implemented as doubly-linked-list) is O(1) for append and prepend, but O(n) for retrieval. Deque (ArrayDeque) is O(1) for everything above but cannot retrieve element at arbitrary index. In my mind a data structure that satisfy the requirement above has 2 growable list inside (one for prepend and one for append) and also stores an offset

android application dequeuebuffer error on snapdragon device

旧时模样 提交于 2019-12-10 01:58:54
问题 I got some strange crash when running our android application on snapdragon device, but thing works well on other devices. Here are some logs before process die: W/Adreno-EGLSUB( 3075): <DequeueBuffer:583>: dequeue native buffer fail: Unknown error 2147483646, buffer=0x0, handle=0x0 W/Adreno-EGLSUB( 3075): <SwapBuffers:1300>: Invalid native buffer. Failed to queueBuffer W/Adreno-EGLSUB( 3075): <updater_thread:456>: native buffer is NULL D/QCUtilClass( 2464): extended extractor not needed,

Will STL deque pop_front() automatically recycle memory?

天大地大妈咪最大 提交于 2019-12-08 07:22:12
问题 I have a program in which I collect some data and store them temporarily in a deque typedef vector<float> floatVector; ... floatVector * currRecord; deque<floatVector *> data; ... ... for (...) { ... currRecord = new floatVector(10); data.push_back(currRecord); } Later, I want to save data to file while (data.size() > 0) { for (int i=0; i < 10; i++) { fprintf(fPtr, "%lf\t", data[0]->at(i) ); } fprintf(fPtr,"\n"); data.pop_front(); } So, my question is, will this program cause a memory leak? I

How to implement a dequeue using two stacks

左心房为你撑大大i 提交于 2019-12-07 18:19:41
问题 Dequeue - Doubly ended queue: en-queue and de-queue possible from both ends. How to define ADT operations for dequeue using 2 stacks. Implementation should also take into consideration the performance. 回答1: the simplest solution would be to use one stack as the head of the queue, and one as the tail. The enqueue operations would just be a push to the respective stack, and the dequeue operations would just be a pop on the respective stack. However, if the stack you want to dequeue from is

deque::insert() at index?

我们两清 提交于 2019-12-07 13:00:46
问题 How do I insert() a bunch of items to the middle of a deque in linear time? (The items I am inserting are not accessible through an STL-style iterator.) 回答1: There is a deque::insert(iterator pos, const T&x) function taking the position pos as deque::iterator and a single element. Using this method you could insert all elements one by one. pos can easily be obtained (if you have an index before which you want to insert the element) by deque.begin()+index . The insert method returns an

ArrayDeque and LinkedBlockingDeque

*爱你&永不变心* 提交于 2019-12-06 06:42:25
问题 Just wondering why they made a LinkedBlockingDeque while the same non concurrent counterpart is an ArrayDeque which is backed on a resizable array. LinkedBlockingQueue use a set of nodes like a LinkedList (even though not implementing List ). I am aware of the possibility to use an ArrayBlockingQueue but what if one wanted to use an ArrayBlockingDeque ? Why is there not such an option? Thanks in advance. 回答1: This May not be a proper question w.r.t stackoverflow. But i would Like to say

Problem with invalidation of STL iterators when calling erase

走远了吗. 提交于 2019-12-06 05:45:48
问题 The STL standard defines that when an erase occurs on containers such as std::deque, std::list etc iterators are invalidated. My question is as follows, assuming the list of integers contained in a std::deque, and a pair of indicies indicating a range of elements in the std::deque, what is the correct way to delete all even elements? So far I have the following, however the problem here is that the assumed end is invalidated after an erase: #include <cstddef> #include <deque> int main() { std

Why does java linkedlist implementation use the interface deque?

瘦欲@ 提交于 2019-12-05 23:39:01
I was looking at the java implementation of LinkedList, and found this: public class LinkedList<E> extends AbstractSequentialList<E> implements List<E>, Deque<E>, Cloneable, java.io.Serializable Why should a LinkedList support the Deque interface? I understand the desire to add elements to the end of the linked list, but those methods should have been incuded in the List interface. The LinkedList implementation happens to to satisfy the Deque contract, so why not make it implement the interface? IIRC, deque stands for double end queue . In the case you mention, it's not logical to define a

How to implement a dequeue using two stacks

一个人想着一个人 提交于 2019-12-05 21:15:51
Dequeue - Doubly ended queue: en-queue and de-queue possible from both ends. How to define ADT operations for dequeue using 2 stacks. Implementation should also take into consideration the performance. the simplest solution would be to use one stack as the head of the queue, and one as the tail. The enqueue operations would just be a push to the respective stack, and the dequeue operations would just be a pop on the respective stack. However, if the stack you want to dequeue from is empty, you'd have to pop each element from the other stack and push it back to the stack you want to dequeue

will stl deque reallocate my elements (c++)?

折月煮酒 提交于 2019-12-05 05:34:52
Hi I need an stl container which can be indexed like a vector but does not move old elements in the memory like a vector would do with resize or reserve (Unless I call reserve once at the beginning with a capacity enough for all elements, which is not good for me). (Note I do address binding to the elements so I expect the address of these elements to never change). So I've found this deque. Do you think it is good for this purpose? Important: I need only pushback but I need to grow the container on demand in small chunks. std::deque "never invalidates pointers or references to the rest of the