recursion

问题 I can't understand this recursion even though it's a really simple example. When it goes to power(base, exponent - 1); what is that supposed to do? How are things being multiplied when power keeps getting invoked until exponent equals 0? function power(base, exponent) { if (exponent === 0) { return 1; } else { return base * power(base, exponent - 1); } } 回答1: Let's start from the beginning. Let's say you call power(base, 0) . Since exponent is 0, the function returns 1. Now, let's say you

问题 The problem I am tackle with is to find the first occurrence node in its inorder traversal in a BST. The code I have is given below def Inorder_search_recursive(node,key): if not node: return None InOrder_search_recursive(node.lChild) if node.value==key: return node InOrder_search_recursive(node.rChild) This code always return None, what's wrong with it. I think I've return node when I find a node with value k. Why cannot python pass this node???Thanks in advance 回答1: When you call yourself

问题 I've written a javascript to emulate WPF DockPanel Layout behaviour in HTML through the help of javascript. Now i'm running into performance issues once i begin nesting those panels to a recursion level of 10. Oddly enough it's nothing more than ordinary recursion and on the deepest level the function in question is finishing in between 0,2 and 2ms. Now either i do have some ghost performance loss, or is there a massive cost in invoking recursion for javascript? I hope one of you knows. If

问题 As we all know, the simplest algorithm to generate Fibonacci sequence is as follows: if(n<=0) return 0; else if(n==1) return 1; f(n) = f(n-1) + f(n-2); But this algorithm has some repetitive calculation. For example, if you calculate f(5), it will calculate f(4) and f(3). When you calculate f(4), it will again calculate both f(3) and f(2). Could someone give me a more time-efficient recursive algorithm? 回答1: One simple way is to calculate it iteratively instead of recursively. This will

问题 I made a function that searches files recursively and I want it to stop recursion when the first file is found: def search_file(path): for name in os.listdir(path): sub = os.path.join(path, name) if os.path.isfile(sub): return sub#And break recursion else: search_file(sub) 回答1: Return a flag that says whether the file was found. When you call search_file , return if the return value is True. 回答2: You are close. You already break recursion when you find the file, the problem is that you didn't

问题 I have the array [1, 2, 4] and I want to make 4; I want it to return [4], not [[1,1,1,1], [1,1,2], [2,2], [4]], I was previously doing this but I started to run out of memory when running this on a small server. I had to let the whole function run through so I could .reverse() because the last time in the array was only useful to me. If we think of this in terms of currency, let's take [0.5, 0.5, 1, 0.2]; if I want $1, I want [1] and not [0.5, 0.5] because you always want the largest value

问题 I have the array [1, 2, 4] and I want to make 4; I want it to return [4], not [[1,1,1,1], [1,1,2], [2,2], [4]], I was previously doing this but I started to run out of memory when running this on a small server. I had to let the whole function run through so I could .reverse() because the last time in the array was only useful to me. If we think of this in terms of currency, let's take [0.5, 0.5, 1, 0.2]; if I want $1, I want [1] and not [0.5, 0.5] because you always want the largest value

问题 I'm trying to write a function with the type declaration [(Int, Bool)] -> [[Int]] . I want the function to only add Int s to the same nested sublist if the Boolean is True . However if the Boolean is False , I want the Int associated with the next True bool to be added to a new sublist. For example: An input of [(1,True),(2,True),(3,False),(4,True),(5,False),(6,False),(7,True)] should return [[1,2],[4],[7]]. My code so far: test:: [(Int, Bool)] -> [[Int]] test xs = case xs of []->[] x:xs |

问题 void MergeSort(int A[], int n, int B[], int C[]) { if(n > 1) { Copy(A,0,floor(n/2),B,0,floor(n/2)); Copy(A,floor(n/2),n-1,C,0,floor(n/2)-1); MergeSort(B,floor(n/2),B,C); MergeSort(C,floor(n/2),B,C); Merge(A,B,0,floor(n/2),C,0,floor(n/2)-1); } }; void Copy(int A[], int startIndexA, int endIndexA, int B[], int startIndexB, int endIndexB) { while(startIndexA < endIndexA && startIndexB < endIndexB) { B[startIndexB]=A[startIndexA]; startIndexA++; startIndexB++; } }; void Merge(int A[], int B[],int

问题 I have to create a recursive Scheme function for my programming class that will take all the odd-numbered elements of a list, and then return them in reversed order. I have a function for reversing a list, and another function for getting the odd-numbered elements, but can't figure out how to combine the two into a new function, as they both are recursive. It has to be one function that doesn't call any functions other than itself. It can not call odd or reverse , and it has to have the same