queue

问题 Can we pause the MDB message processing for some time? For example: Jboss 1-deployed MDB for message processing. Jboss 2:-Bean for gathering user details. If the MDB from jboss 1 calls bean in jboss 2 for getting users details. If this is the case, when we restart the Jboss 2, we need to pause the MDB in jboss 1 till the jboss 2 is UP. Is there any option to pause MDB, so that we can avoid failure of message? 回答1: I doubt you can stop an MDB without stopping the whole application. It is

问题 I have a Laravel based application which is image intensive. Users can upload images to the server and the images are stored on Amazon s3 bucket after being resized. The process is pretty slow here and I've been reading up on queues and think they may be exactly what i need to kind of delegate the part of storing on amazon to. The only thing is that this is my postAction which handles the uploading: public function postImage(){ $images = Input::only('images'); $model->saveImages($images[

问题 I have a Laravel based application which is image intensive. Users can upload images to the server and the images are stored on Amazon s3 bucket after being resized. The process is pretty slow here and I've been reading up on queues and think they may be exactly what i need to kind of delegate the part of storing on amazon to. The only thing is that this is my postAction which handles the uploading: public function postImage(){ $images = Input::only('images'); $model->saveImages($images[

问题 I have a queue with some numbers for example 5,6,7,8 is in queue, 5 is q->front and 8 is q->rear.. Can I reverse them in queue but without using stacks? 回答1: Of course! But it won't be as efficient. Using a stack: O(N) time. O(1) memory. Using an associative array: O(N) time. O(1) memory. Using a fixed-size array: O(N) time. O(N) memory. Using an extendable array: O(N) time. O(N) memory. Using a queue: O(N^2) time. O(1) memory. Using an associative array will use more time than using a stack,

问题 I have a queue with some numbers for example 5,6,7,8 is in queue, 5 is q->front and 8 is q->rear.. Can I reverse them in queue but without using stacks? 回答1: Of course! But it won't be as efficient. Using a stack: O(N) time. O(1) memory. Using an associative array: O(N) time. O(1) memory. Using a fixed-size array: O(N) time. O(N) memory. Using an extendable array: O(N) time. O(N) memory. Using a queue: O(N^2) time. O(1) memory. Using an associative array will use more time than using a stack,

问题 My application needs a data structure similar to a queue: put data in it, receive data from it, FIFO-like. With data I mean simple strings for now, later on perhaps more complex objects. The thing is that the queue and its content should be persistent, regardless of what Android is doing. If the application gets closed and reopened (or even Android reboots), the queue should have the same state and data it had before the application was closed. I think the queue has to use some kind of

问题 Which one of Queue's subclasses is a 'plain ordinary' queue? 回答1: (1) java.util.Stack is a legacy class from Java 1.0. It predates the Collections framework by many years, and it's frankly an example of horrible design on many fronts. Nothing about it is the way things should be. The main problem is that Stack extends Vector , and as all inheritance in Java is public inheritance, all the methods of Vector are available on Stack as well. Therefore, you can inspect any position in a Stack, add

问题 I was working on a problem on LeetCode (Here). When I finished the problem, I came up with: class MovingAverage { std::deque<int> numsToAverage; int maxSize; int currentTotal; public: /** Initialize your data structure here. */ MovingAverage(int size) { maxSize = size; currentTotal = 0; } double next(int val) { currentTotal += val; numsToAverage.push_back(val); if (numsToAverage.size() > maxSize) { currentTotal -= numsToAverage[0]; numsToAverage.pop_front(); } return (double)currentTotal /

问题 I was working on a problem on LeetCode (Here). When I finished the problem, I came up with: class MovingAverage { std::deque<int> numsToAverage; int maxSize; int currentTotal; public: /** Initialize your data structure here. */ MovingAverage(int size) { maxSize = size; currentTotal = 0; } double next(int val) { currentTotal += val; numsToAverage.push_back(val); if (numsToAverage.size() > maxSize) { currentTotal -= numsToAverage[0]; numsToAverage.pop_front(); } return (double)currentTotal /

问题 I am having trouble understanding how the deque works in the snippet of code below, while trying to recreate a queue and a stack in Python. Stack Example - Understood stack = ["a", "b", "c"] # push operation stack.append("e") print(stack) # pop operation stack.pop() print(stack) As expected when pushing and popping, the "e" goes Last In, First Out (LIFO). My question is with the example below. Queue Example - Not Understanding from collections import deque dq = deque(['a','b','c']) print(dq)