recursion

问题 I'm trying to write a function that reverses a list. The function is recursive. I know javascript does not have TCO, but i wanted to experiment with this anyway: reverse = function(list) { if (list.length < 2) { return list } fk = fork(list); return reverse(fk.tail).concat([fk.head]) } the fork function splits a list to a head and a tail: fork = function(list) {return {head: list[0], tail: list.slice(1)}} When I call reverse() with the list [1,2,3,4,5] , I get this result: reverse([1,2,3,4,5]

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 6 years ago . This is a sample program for my problem, I am using VisualStudio 2008 void abc() { static int i = 0; if (i==0) { xyz(); i++; } abc(); } The static variable retain the value one in next debug session also, thus not calling xyz() , how can I call a function just once without using static variable??

问题 I have a category tree array fetched from MySQL Table. I want to revert this Category array tree back into Breadcrumb list using PHP. PHP Category Tree Building Function function buildTree(array &$elements, $parentId = 0) { $branch = array(); foreach ($elements as $element) { if ($element['parent_category_id'] == $parentId) { $children = buildTree($elements, $element['category_id']); if ($children) { $element['children'] = $children; } $branch[$element['category_id']] = $element; unset(

问题 What's the best way in Python to recursively go through all directories until you find a certain file? I want to look through all the files in my directory and see if the file I'm looking for is in that directory. If I cannot find it, I go to the parent directory and repeat the process. I also want to count how many directories and files I go through before I find the file. If there is no file at the end of the loop return that there is no file startdir = "Users/..../file.txt" findfile is

问题 What's the best way in Python to recursively go through all directories until you find a certain file? I want to look through all the files in my directory and see if the file I'm looking for is in that directory. If I cannot find it, I go to the parent directory and repeat the process. I also want to count how many directories and files I go through before I find the file. If there is no file at the end of the loop return that there is no file startdir = "Users/..../file.txt" findfile is

问题 I have a problem with a @Get method. I have an entity ServcieChargeTier which has a @OneToMany relationship with the entity CalendarEntry . The problem is when I try and get a ServiceChargeTier from the server, the server returns a recursive loop of the ServiceChargeTier, which has CalendarEntries , which each has a ServiceChargeTier associated, which has the CalendarEntries and so on. I would like to return the CalendarEntry's but not the associated ServiceChargeTier for each CalendarEntry.

问题 The original question is here: Cartesian power (a special Cartesian product) -- choose elements from array, in repeatable style In the old question, there are already answers gave a solution via iteration. I am wondering is there a recursive solution, similar as the solution from following link, which print permutations with recursion: https://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/ Currently I have wrote following program, which is not correct yet

问题 The original question is here: Cartesian power (a special Cartesian product) -- choose elements from array, in repeatable style In the old question, there are already answers gave a solution via iteration. I am wondering is there a recursive solution, similar as the solution from following link, which print permutations with recursion: https://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/ Currently I have wrote following program, which is not correct yet

问题 So i've managed to get Pascals Triangle to print successfully in terms of what numbers are printed, however, i can't get the formatting correct using: n = int(input("Enter value of n: ")) def printPascal(n): if n <= 0: #must be positive int return "N must be greater than 0" elif n == 1: #first row is 1, so if only 1 line is wanted, output always 1 return [[1]] else: next_row = [1] #each line begins with 1 outcome = printPascal(n-1) prev_row = outcome[-1] for i in range(len(prev_row)-1): #-1

问题 My understanding of continuous fractions was that it will always give a representation of a decimal in fraction form. I thought that continuous fraction would always return value less than or equal to the decimal number. Unfortunately my code occasionally returns fractional values greater than the decimal input. Is my understanding of continuous fractions correct? If so can you please explain where in my code the error lies. public static Rational contFrac(double a, int i,int n){ if(i<n){