如何将数组推送到递归 javascript 函数内的数组，以便在完成后可以使用它？

Question

First question here (I think). Please let me know if additional information is needed in order for you guys to help me out.

So I'm trying to implement an algorithm in javascript that uses a recursive funtion.

The function is copied from Implementing Heap Algorithm of Permutation in JavaScript and looks like this:

let swap = function(array, index1, index2) {
    let temp = array[index1]
    array[index1] = array[index2]
    array[index2] = temp
    return array
}

let permutationHeap = (array, result, n) => {
    n = n || array.length // set n default to array.length
    if (n === 1) {
        result(array)
    } else {
        for (let i = 1; i <= n; i++) {
            permutationHeap(array, result, n - 1)
            if (n % 2) {
                swap(array, 0, n - 1) // when length is odd so n % 2 is 1,  select the first number, then the second number, then the third number. . . to be swapped with the last number
            } else {
                swap(array, i - 1, n - 1) // when length is even so n % 2 is 0,  always select the first number with the last number
            }
        }
    }
}


let output = function(input) {
    console.log(output)
}

permutationHeap([1,2,3,4,5], output)

The console.log in the output function (callback?) gives me the right output. If I move that console.log below the if statement in permutationHeap-function, I get the right output as well (console.log(array), in that case though).

What I want to do is to store every output as an array, inside an array that I can use later on down the road. I'm guessing that I'm struggling with Javascript 101 here. Dealing with asynchronus thinking. But can't for the life of me figure out how to get that array of arrays!

• If I declare an empty array outside the permutationHeap-function and .push(array) it only stores [1,2,3,4,5]. Same deal if I do the same thing inside the output-function.
• I've also tried passing in an empty array to the permutationHeap-function and push that way. Still no luck.

Anyone who's willing to shine some light over a probably super nooby question? :) Much appriciated!

Answer 1

I think I've managed to fix your code by using generators and yield. I can't exactly explain it though...

 var swap = function(array, index1, index2) { let temp = array[index1] array[index1] = array[index2] array[index2] = temp return array } var permutationHeap = function*(array, result, n) { n = n || array.length // set n default to array.length if (n === 1) { yield (array.slice(0)) } else { for (let i = 1; i <= n; i++) { yield* permutationHeap(array, result, n - 1) if (n % 2) { swap(array, 0, n - 1) // when length is odd so n % 2 is 1, select the first number, then the second number, then the third number. . . to be swapped with the last number } else { swap(array, i - 1, n - 1) // when length is even so n % 2 is 0, always select the first number with the last number } } } } var x = permutationHeap([1,2,3,4,5]) var results = Array.from(x); console.log(results);

Answer 2

I actually broke the algorithm in my previous answer, because by duplicating the array ever iteration I broke the part of the algorithm that relies on the array changing as it goes.

I've made an answer that uses the code that you had originally more effectively:

 var swap = function(array, index1, index2) { var temp = array[index1]; array[index1] = array[index2]; array[index2] = temp; return array; }; var permutationHeap = function(array, result, n) { n = n || array.length; // set n default to array.length if (n === 1) { result(array); } else { for (var i = 1; i <= n; i++) { permutationHeap(array, result, n - 1); if (n % 2) { swap(array, 0, n - 1); // when length is odd so n % 2 is 1, select the first number, then the second number, then the third number. . . to be swapped with the last number } else { swap(array, i - 1, n - 1); // when length is even so n % 2 is 0, always select the first number with the last number } } } }; function getPermutations(array) { var results = []; var output = function(res) { results.push(res.slice(0)); } permutationHeap(array, output); return results; } var permutations = getPermutations([1,2,3]); console.log(permutations);