recursion

Recursive function returns undefined regardless of enough return statements

有些话、适合烂在心里 提交于 2021-01-27 14:50:26
问题 I have read a few questions and answers on it already. It looks like my recursive function has got enough "return" statements, so... I do not know why it returns undefined... I have added extra log statement to show that the function itself finds the element, but does not return it... let animals = [ { name: "dogs", id: 1, children: [ { name: "lessie", id: 2 }, { name: "bark-a-lot", id: 3 } ] }, { name: "cats", id: 4, children: [ { name: "meows-a-lot", id: 5, children: [ { name: "meows-a-lot

How to recursively add items to a list?

我只是一个虾纸丫 提交于 2021-01-27 14:15:18
问题 Currently, I'm working on a problem. I am given a list, whose elements may contain other lists, lists of lists, or integers. For example, I may receive: [[[[], 1, []], 2, [[], 3, []]], 4, [[[], 5, []], 6, [[], 7, [[], 9, []]]]] My goal is to parse the array, and append only the integers to a new list. Here is what I have done so far: def fun(a): if a == []: return None elif type(a) == int: print("Found a digit: ", a) return a for i in a: fun(i) Currently, this function recursively goes

Contradiction about Order of Evaluation of operands

蹲街弑〆低调 提交于 2021-01-27 07:24:57
问题 When I study recursive functions in C from deitel c, I read this sentence: Standard C does not specify the order in which the operands of most operators (including +) are to be evaluated. But also the book says that: associativity of '+' from left to right Order of evaluation of operands: Could anyone explain why this is? 回答1: Order of evaluation and associativity are two different things, take the example: int x = func1() - func2() - func3(); //having int return types In this expression you

Can this implementation of Ackermann function be called tail recursive?

不打扰是莪最后的温柔 提交于 2021-01-27 07:20:14
问题 I have written following code in C. Can we call it a tail recursive implementation? #include <stdio.h> int ackermann(unsigned int *m, unsigned int *n, unsigned int* a, int* len) { if(!*m && *len == -1) { return ++*n; } else if(!*m && *len >= 0) { ++*n; *m = a[(*len)--]; } else if(*n == 0) { --*m; *n = 1; } else { ++*len; a[*len] = *m - 1; --*n; } return ackermann(m, n, a, len); } int main() { unsigned int m=4, n=1; unsigned int a[66000]; int len = -1; for (m = 0; m <= 4; m++) for (n = 0; n <

String permutations using recursion in Java

六月ゝ 毕业季﹏ 提交于 2021-01-27 06:45:31
问题 I came across THIS post which tried very hard to explain the recursive solution to print all string. public class Main { private static void permutation(String prefix, String str){ int n = str.length(); if (n == 0) System.out.println(prefix); else { for (int i = 0; i < n; i++) permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1)); } } public static void main(String[] args) { permutation("", "ABCD"); } } But still I am not able to get the part when we start popping off

Methods to exhaustively partition a vector into pairs in R

大憨熊 提交于 2021-01-27 06:23:53
问题 (This is inspired by another question marked as a duplicate. I think it is an interesting problem though, although perhaps there is an easy solution from combinatorics, about which I am very ignorant.) Problem For a vector of length n , where n mod 2 is zero, find all possible ways to partition all elements of the vector into pairs, without replacement, where order does not matter. For example, for a vector c(1,2,3,4) : list(c(1,2), c(3,4)) list(c(1,3), c(2,4)) list(c(1,4), c(2,3)) My

Methods to exhaustively partition a vector into pairs in R

泄露秘密 提交于 2021-01-27 06:23:08
问题 (This is inspired by another question marked as a duplicate. I think it is an interesting problem though, although perhaps there is an easy solution from combinatorics, about which I am very ignorant.) Problem For a vector of length n , where n mod 2 is zero, find all possible ways to partition all elements of the vector into pairs, without replacement, where order does not matter. For example, for a vector c(1,2,3,4) : list(c(1,2), c(3,4)) list(c(1,3), c(2,4)) list(c(1,4), c(2,3)) My

ANTLR4 mutually left-recursive error when parsing

非 Y 不嫁゛ 提交于 2021-01-21 12:10:06
问题 I have this ANTLR 4 grammar: constantFixedExpresion : term (('+'|'-') term)+; term : factor (('*'|'//'|'REM')factor)+; factor : ('+'|'-')* ( wholeNumberConstant | constantFixedExpresion | 'TOFIXED' (stringConstant | bitCodeConstant) | identifier) ('FIT'constantFixedExpresion)*; I get the following error: error(119): LanguageA.g4::: The following sets of rules are mutually left-recursive [constantFixedExpresion, factor, term] I tried so many ways but can't fix it. What is the problem and how

ANTLR4 mutually left-recursive error when parsing

穿精又带淫゛_ 提交于 2021-01-21 12:08:13
问题 I have this ANTLR 4 grammar: constantFixedExpresion : term (('+'|'-') term)+; term : factor (('*'|'//'|'REM')factor)+; factor : ('+'|'-')* ( wholeNumberConstant | constantFixedExpresion | 'TOFIXED' (stringConstant | bitCodeConstant) | identifier) ('FIT'constantFixedExpresion)*; I get the following error: error(119): LanguageA.g4::: The following sets of rules are mutually left-recursive [constantFixedExpresion, factor, term] I tried so many ways but can't fix it. What is the problem and how

C# parse recursive json

南笙酒味 提交于 2021-01-21 11:47:09
问题 I am trying to parse a JSON string into a C# class. however i can't get it to work. the JSON looks like this { reports: { report: { id:"1" step: { id:"2" step: [ { id:"3" step: [ { id:"4" }, { id:"5" } ] }, { id:"6" step: { id:"7" } } ] } } } } i made a few classes to define the structure. private class report { public mongoReports reports; } private class mongoReports { public mongoReport report; } private class mongoReport { public string id; public step step; } private class step { public