sorting

问题 I have a Class called apple which contains 3 values as int x , int y and int weight . Then i created an array of apple type objects. Now i want to sort the the array of objects based on weight meaning the the apple object with the lowest weight should be first and so on. I know there are quite a few ways to achieve this by using Arrays.sort etc or comparators. I was wondering what is the fastest way of doing this sort in Java? There can be a case where i have 500,000 objects so i want to know

问题 I have a dictionary that looks like this: myDict = { 'SER12346': {'serial_num': 'SER12346', 'site_location': 'North America'}, 'ABC12345': {'serial_num': 'ABC12345', 'site_location': 'South America'}, 'SER12345': {'serial_num': 'SER12345', 'site_location': 'North America'}, 'SER12347': {'serial_num': 'SER12347', 'site_location': 'South America'}, 'ABC12346': {'serial_num': 'ABC12346', 'site_location': 'Europe'} } My goal is to sort this dictionary by the site_location and the serial_num of

问题 I'm building an app using Swift with Core Data. At one point in my app, I want to have a UITableView show all the objects of a type currently in the Persistent Store. Currently, I'm retrieving them and showing them in the table view using an NSFetchedResultsController . I want the table view to be sorted by a computed property of my NSManagedObject subclass, which looks like this: class MHClub: NSManagedObject{ @NSManaged var name: String @NSManaged var shots: NSSet var averageDistance: Int{

问题 I am working on a Google Apps Script spreadsheet application, and one of the abilities I would like the program to have is to automatically sort a series of form responses based on data from 2 different columns. So I would want to sort it by the data in column 16 and then sort by column 1. I can achieve this functionality manually using the method at: https://drive.googleblog.com/2010/06/tips-tricks-advanced-sorting-rules-in.html Currently I am running the Spreadsheet.sort(column, ascending)

问题 Probably best illustrated with a small example. Given the relations A < B < C A < P < Q Correct outputs would be ABCPQ or APQBC or APBCQ ... etc. In other words, any ordering is valid in which the given relationships hold. I am most interested in the solution that is easiest to implement, but the best O(n) in speed and time is interesting as well. 回答1: This is called topological sorting. The standard algorithm is to output a minimal element, then remove it and repeat until done. 回答2: Do

问题 Probably best illustrated with a small example. Given the relations A < B < C A < P < Q Correct outputs would be ABCPQ or APQBC or APBCQ ... etc. In other words, any ordering is valid in which the given relationships hold. I am most interested in the solution that is easiest to implement, but the best O(n) in speed and time is interesting as well. 回答1: This is called topological sorting. The standard algorithm is to output a minimal element, then remove it and repeat until done. 回答2: Do

问题 I have a combo box that triggers a jquery function on change, it calls a php script which brings the results from the database sorted by name not by id (table cities) and creates a json file with the id and the name of the city. The problem comes with a $.getJSON(url, data) function, it retrieves all the json encoded data fine, but it is seems to be sorting the data automatically by the id for instance if the php page generates id name 3 Dania Beach 1 Miami 2 Weston after jquery getJSON it

问题 I have a list of dictionaries: [{'title':'New York Times', 'title_url':'New_York_Times','id':4}, {'title':'USA Today','title_url':'USA_Today','id':6}, {'title':'Apple News','title_url':'Apple_News','id':2}] I'd like to sort it by the title, so elements with A go before Z: [{'title':'Apple News','title_url':'Apple_News','id':2}, {'title':'New York Times', 'title_url':'New_York_Times','id':4}, {'title':'USA Today','title_url':'USA_Today','id':6}] What's the best way to do this? Also, is there a

问题 Given n numbers, how do I find the largest and second largest number using at most n+log(n) comparisons? Note that it's not O(n+log(n)), but really n+log(n) comparisons. 回答1: pajton gave a comment. Let me elaborate. As pajton said, this can be done by tournament selection. Think of this as a knock out singles tennis tournament, where player abilities have a strict order and the outcome of a match is decided solely by that order. In the first round half the people are eliminated. In the next

问题 Given n numbers, how do I find the largest and second largest number using at most n+log(n) comparisons? Note that it's not O(n+log(n)), but really n+log(n) comparisons. 回答1: pajton gave a comment. Let me elaborate. As pajton said, this can be done by tournament selection. Think of this as a knock out singles tennis tournament, where player abilities have a strict order and the outcome of a match is decided solely by that order. In the first round half the people are eliminated. In the next