如何获得一个数组，其中包含另一个数组的 n 个元素的所有可能排列？

Question

I started studying not so long ago [this is actually my first post] but I decided to try a little side project in javascript and I encountered this challenge.

I have an array with 7 elements right now, but i plan to change its length in the future. I need another array with all the possible combinations of N elements of the first array.

So far I could only find a way to get all combinations (N = firstArray.length-1), but haven't had any luck with a lower N.

Can anyone shed some light? Thanks!

Answer 1

Here is one way to get the answer. To understand the idea, begin by imagining you want to find all arrangements of an array of two, non-like elements [A,B]. Do not think of this array as [A,B]. Think of it as [0,1]. Since all combinations of these two elements can be found by counting from 0 to 3 in base 2, we have [0,0],[0,1],[1,0] and [1,1]. So counting from 1 to 3 in base two will give you all possible combinations, and if you throw out all the arrays with repeat digits and then substitute back with A for 0 and B for 1, you have your answer.

So to solve your seven-element problem,

1. Declare a variable, say "let allCombinations = []; to store all your possible combinations.

2. Work in base seven and make the computer count from [0,0,0,0,0,0,0] to [6,6,6,6,6,6,6] by base seven.

3. Check each new base-seven array for any repeat numbers using a loop from 0 to six to check each element in the array to see if it occurs more than once.

4. If elements occur more than once in the current array, toss it out. If not, push it to allCombintation.

5. Continue this process until you reach [6,6,6,6,6,6,6]. Then map through the arrays and replace each digit with its corresponding original element.

This process can be modified to account for sets with repeat element as well. But the basic problem this approach solves is ensuring every possible arrangement of elements is accounted for without missing any arrangements.