recursion

问题 I'm currently working with a recursive function in Python, and I've run into a wall. As titled, the problem is to return the maximum depth of an arbitrarily nested list. Here is what I have so far: def depthCount(lst): 'takes an arbitrarily nested list as a parameter and returns the maximum depth to which the list has nested sub-lists.' var = 0 if len(lst) > 0: if type(lst[0]) == list: var += 1 depthCount(lst[1:]) else: depthCount(lst[1:]) else: return var I feel that the problem is with my

问题 I'm currently working with a recursive function in Python, and I've run into a wall. As titled, the problem is to return the maximum depth of an arbitrarily nested list. Here is what I have so far: def depthCount(lst): 'takes an arbitrarily nested list as a parameter and returns the maximum depth to which the list has nested sub-lists.' var = 0 if len(lst) > 0: if type(lst[0]) == list: var += 1 depthCount(lst[1:]) else: depthCount(lst[1:]) else: return var I feel that the problem is with my

问题 I am currently trying to make a script that will find images with *.jpg / *.png extensions in directories and subdirectories. If some picture with one of these extensions is found, then save it to an array with path, name, size, height and width. So far I have this piece of code, which will find all files, but I don't know how to get only jpg / png images. class ImageCheck { public static function getDirectory( $path = '.', $level = 0 ){ $ignore = array( 'cgi-bin', '.', '..' ); // Directories

问题 I am currently trying to make a script that will find images with *.jpg / *.png extensions in directories and subdirectories. If some picture with one of these extensions is found, then save it to an array with path, name, size, height and width. So far I have this piece of code, which will find all files, but I don't know how to get only jpg / png images. class ImageCheck { public static function getDirectory( $path = '.', $level = 0 ){ $ignore = array( 'cgi-bin', '.', '..' ); // Directories

问题 For example given ['a', 'b'] (as a generator) and 2 as a length the function would output a generator that would yield: '', 'a', 'b', 'ab' 'ba' 'aa' 'bb' or given ['a'] and a length of 3: '', 'a', 'aa', 'aaa', As you could imagine this set would get a lot larger if more letters were added or length was increased, it should list all permutations of the given characters up until the length 回答1: Here's a fairly self-explanatory solution. //Returns all permuations of a certain length. function

问题 I have got a flat array representing a tree, and I want to build a nested object using tail recursion. I've got the following code to run and generate the desired output, but I am not sure if it is a proper implementation of tail recursion. Please advice :) const myArray = [ { id: 'root' }, { id: 0, parent: 'root' }, { id: 1, parent: 'root' }, { id: 2, parent: 0 }, { id: 3, parent: 1 }, { id: 4, parent: 2 }, { id: 5, parent: 1 }, { id: 6, parent: 4 }, { id: 7, parent: 0 }, { id: 8, parent: 0

问题 I want to write a type utility that omits fields recursively. Something that you would name and use like that OmitRecursively<SomeType, 'keyToOmit'> I've tried to do it using mapped types + conditional typing but I stuck on the case when all required fields got typed correctly (hence field disappeared from nested type), but optional fields are ignored with that approach. // This is for one function that removes recursively __typename field // that Appolo client adds type Deapolify<T extends {

问题 I have the following Jinja2 template: {% block body %} {% for key in tree recursive %} {% set outer_loop = loop %} {% for subkey in tree[key] %} {% if subkey == 'R' %} {{ tree[key][subkey] }} {% else %} {{ outer_loop(dict([(subkey, tree[key][subkey])])) }} {% endif %} {% endfor %} {% endfor %} {% endblock body %} where tree is a Python dictionary such as: tree = {"A": {"R": [1, 2, 3], "B": {"R": [4, 5, 6]}}} and dict() is the Python library function. The issue is that the template displays

问题 I want to find the number of characters inside a double quotation mark. For example : Case 1 "Hello World" , "Some output : Error // missing quotation marks after Some Case 2 "Hello Word" , "Some" output : 14 // All quotation marks are complete I have written the program for counting the total characters, the index of the first quotation mark and the total number of quotations marks using recursion. What approach shall I use to solve this problem? 回答1: Here is a working version using

问题 So I have this code which I found on the internet which takes an array of negative and positive numbers and rearranges the array so all negative numbers are before the positive ones. But the position of appearance of each number has to remain the same. So for example if i have -2 5 -9 ..... , in the organized array -2 still has to be the first number of the negative ones and -9 has to be the second of the negative ones, also 5 has to be the first number of the positive ones. Example: Input: