permutation | 易学教程

How to convert this function from recursive to iterative?

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问题 I have this method: static List<string> permutation(List<string> lst) { switch (lst.Count) { case 0: case 1: return lst; default: List<string> l = new List<string>(); for (int i = 0; i < lst.Count; i++) { string m = lst[i]; List<string> remLst = lst.Except(new List<string> { m }).ToList(); List<string> tmp = permutation(remLst); for (int i2 = 0; i2 < tmp.Count; i2++) { l.Add(m + tmp[i2]); } } return l; } } which gives all permutations of the list lst . The idea is to one by one extract all

permutations in Javascript

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问题 I'm trying to write a function in Javascript that can return the number of permutations, and also show all the permutations of a string (suppose that non of the character is repeated) using recursive methods. I've seen a lot using for loop, but is there a way that I can obtain the same outcome without using it? For the number of permutations, here is my attempt without using for loop var permutation = function (s) { var fac = function (t) { if (t === 0) return 1; return t*fac(t-1); }; return

permutations in Javascript

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Permutations with repetition without two consecutive equal elements

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问题 I need a function that generates all the permutation with repetition of an iterable with the clause that two consecutive elements must be different; for example f([0,1],3).sort()==[(0,1,0),(1,0,1)] #or f([0,1],3).sort()==[[0,1,0],[1,0,1]] #I don't need the elements in the list to be sorted. #the elements of the return can be tuples or lists, it doesn't change anything Unfortunatly itertools.permutation doesn't work for what I need (each element in the iterable is present once or no times in

python实现克莱姆法则

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文章目录首先完成python模拟行列式运算公式分析模块分析与实现环境模块导入全排列逆序数方阵计算克莱姆法则 *Cramer's rule* 注：本文对numpy对象使用append方法时均使用了深拷贝deepcopy，因为python中对象的赋值是按引用传递的，如果不使用深拷贝在append时会改变原有对象从而覆盖原先的值首先完成python模拟行列式运算 ∣ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ∣ = ∑ ( − 1 ) t a 1 p 1 a 2 p 2 ⋯ a n p n \begin{vmatrix} {a_{11}}&{a_{12}}&{\cdots}&{a_1n}\\ {a_{21}}&{a_{22}}&{\cdots}&{a_2n}\\ {\vdots}&{\vdots}&{\ddots}&{\vdots}\\ {a_{n1}}&{a_{n2}}&{\cdots}&{a_{nn}}\\ \end{vmatrix}= \sum{(-1)^{t}}{a_{1p_{1}}a_{2p_{2}}}{\cdots}{a_{np_{n}}} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 1 1 a 2 1 ⋮ a n 1 a 1 2 a 2 2 ⋮

Combinations of expressions with 4 elementary operations

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问题 I could not come up with a better title, for an adequate one might require the whole explanation. Also, combinations could be misleading since the problem will involve permutations. What I want to accomplish is to outperform a brute force approach in Python at the following problem: Given the 4 elementary operations [+,-,*,/] and the digits from 1 to 9, and given all the possible combinations of 5 digits and the 4 operations without repetition (permutations) that result in a given number

JavaScript - Generating combinations from n arrays with m elements

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问题 I'm having trouble coming up with code to generate combinations from n number of arrays with m number of elements in them, in JavaScript. I've seen similar questions about this for other languages, but the answers incorporate syntactic or library magic that I'm unsure how to translate. Consider this data: [[0,1], [0,1,2,3], [0,1,2]] 3 arrays, with a different number of elements in them. What I want to do is get all combinations by combining an item from each array. For example: 0,0,0 // item

all permutations of string without using itertools

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问题 All possible strings of any length that can be formed from a given string Input: abc Output: a b c abc ab ac bc bac bca cb ca ba cab cba acb I have tried using this but it's limited to string abc, I want to generalize it like if input 'abcd' i will provide me output for the same. def perm_main(elems): perm=[] for c in elems: perm.append(c) for i in range(len(elems)): for j in range(len(elems)): if perm[i]!= elems[j]: perm.append(perm[i]+elems[j]) level=[elems[0]] for i in range(1,len(elems)):

Combination of a Collection with Repetitions

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问题 There are a lot of links on http://stackoverflow.com for how to do combinations: Generating combinations in c++ But these links presume to draw from an infinite set without repetition. When given a finite collection which does have repetition, these algorithms construct duplicates. For example you can see the accepted solution to the linked question failing on a test case I constructed here: http://ideone.com/M7CyFc Given the input set: vector<int> row {40, 40, 40, 50, 50, 60, 100}; I expect

Combination of a Collection with Repetitions

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