recursion

问题 I want to make a recursive algorithm that solves the change-making problem. Is it possible to use a non-dynamic approach that not only returns the minimum number of coins but also returns the set of coins used to make-up the given value, For example, given the value 6 and the set of coins=[1, 3, 4]. Is it possible to make a recursive algorithm that doesn't memoise that can return both the minimum number of coins (2) and the set of coins (3,3)? EDIT: This is my current algorithm but it only

问题 B B B B B B B B O B B B S O B B O O B B B B X B B Here, S = Start Point(2,2) B = Block O = Open X = Exit I want to make a maze that will check north, west, east and south. if X is around it will return the program. If not, then check for any 'O' around the start point and pass the new start point recursively. It there is no way to go and 'X' is not found it will go back to the original start point(2,2) and check west, east and south. after the program, i got: B B B B B B B B + B B B + + B B O

问题 Out of interest and as a thought exercise, I was aiming to replicate the tree function from the windows command line in R. tree generates ASCII trees similar to the following example (taken from here and modified). If n is the hierarchical level, note that each file or folder has (n-1)*3 spaces, followed by +--- +---Desktop +---Documents +---Custom Office Templates +---Fiddler2 +---Captures +---Requests +---Responses +---Scripts +---Favorites +---Links I use list.files() to extract the

问题 I'm trying to print all the "keys, values" from a json response without knowing the keys names (without using the syntax json['example'], for example). I'm doing this with a recursively function that uses iteritems(), but I'm having some problems: This is the Json response that I'm trying to read: {"servers": [{"id": "a059eccb-d929-43b2-8db3-b32b6201d60f", "links": [{"href": "http://192.168.100.142:8774/v2/2ad1fc162c254e59bea043560b7f73cb/servers/a059eccb-d929-43b2-8db3-b32b6201d60f", "rel":

问题 Problem statement I want to get all possible combinations out of my list (including the empty list). My code so far is: def combination(l): result = [] for item in range(len(l)): cut_list = l[:item] + l[item + 1:] if len(cut_list) > 1: combination(cut_list) elif len(cut_list) == 1: result += cut_list return result print(combination([1, 2, 3])) My output is an empty List [] i want this Output: [[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]] I am pretty sure something with my return is

问题 This code is from akka documentation. It impelements an actor using the recommended functional style: import akka.actor.typed.Behavior import akka.actor.typed.scaladsl.ActorContext import akka.actor.typed.scaladsl.Behaviors object Counter { sealed trait Command case object Increment extends Command final case class GetValue(replyTo: ActorRef[Value]) extends Command final case class Value(n: Int) def apply(): Behavior[Command] = counter(0) private def counter(n: Int): Behavior[Command] =

问题 I'm struggling to turn a simple recursive function into a simple iterator. The problem is that the recursive function maintains state in its local variables and call stack -- and to turn this into a rust iterator means basically externalizing all the function state into mutable properties on some custom iterator struct. It's quite a messy endeavor. In a language like javascript or python, yield comes to the rescue. Are there any techniques in Rust to help manage this complexity? Simple

问题 I'm struggling to turn a simple recursive function into a simple iterator. The problem is that the recursive function maintains state in its local variables and call stack -- and to turn this into a rust iterator means basically externalizing all the function state into mutable properties on some custom iterator struct. It's quite a messy endeavor. In a language like javascript or python, yield comes to the rescue. Are there any techniques in Rust to help manage this complexity? Simple

问题 I wrote some code that calculates the determinant of a given nxn matrix using Leibniz formula for determinants. I am trying to figure out it's complexity in O-notation. I think it should be something like: O(n!) * O(n^2) + O(n) = O(n!*n^2) or O((n+2)!) Reasoning: I think O(n!) is the complexity of permutations. and O(n) the complexity of perm_parity, and O(n^2) is the multiplication of n items each iteration. this is my code: def determinant_leibnitz(self): assert self.dim()[0] == self.dim()

问题 I need a SQL query giving me the complete tree path of one item. The tables Looks like this and there is a 1:n relation between MyItem_MyItemId and MyItemMapping_MyItemId. Table MyItem: MyItem_MyItemId | MyItem_Title 1 | Desktop 2 | Workspace 3 | Folder1 4 | Folder2 5 | Folder3 6 | Folder4 ... Table MyItemMapping: MyItemMapping_MyItemId | MyItemMapping_MyItemParentId 4 | 3 3 | 2 2 | 1 1 | NULL 5 | 2 6 | 2 ... Now I Need a query the brings the path for the Folder2 like "Desktop\Workspace