查找JavaScript数组值的所有组合（笛卡尔积） [英] Finding All Combinations (Cartesian product) of JavaScript array values

问题描述

如何在N个可变长度的JavaScript数组中生成值的所有组合？

How can I produce all of the combinations of the values in N number of JavaScript arrays of variable lengths?

假设我有N个JavaScript数组，例如

Let's say I have N number of JavaScript arrays, e.g.

var first = ['a', 'b', 'c', 'd'];
var second = ['e'];
var third =  ['f', 'g', 'h', 'i', 'j'];

（在此示例中为三个数组，但对于问题，使用N个数组）。

(Three arrays in this example, but its N number of arrays for the problem.)

我想输出其值的所有组合，以产生

And I want to output all the combinations of their values, to produce

aef
aeg
aeh
aei
aej
bef
beg
....
dej

---

编辑：这是我使用的版本，使用ffriend接受的答案作为基础

---

Here's the version I got working, using ffriend's accepted answer as the basis.

var allArrays = [['a', 'b'], ['c', 'z'], ['d', 'e', 'f']];

 function allPossibleCases(arr) {
  if (arr.length === 0) {
    return [];
  } 
else if (arr.length ===1){
return arr[0];
}
else {
    var result = [];
    var allCasesOfRest = allPossibleCases(arr.slice(1));  // recur with the rest of array
    for (var c in allCasesOfRest) {
      for (var i = 0; i < arr[0].length; i++) {
        result.push(arr[0][i] + allCasesOfRest[c]);
      }
    }
    return result;
  }

}
var r=allPossibleCases(allArrays);
 //outputs ["acd", "bcd", "azd", "bzd", "ace", "bce", "aze", "bze", "acf", "bcf", "azf", "bzf"]

推荐答案

这不是排列，请参阅Wikipedia中的排列定义。

This is not permutations, see permutations definitions from Wikipedia.

但是您可以通过递归来实现：

But you can achieve this with recursion:

var allArrays = [['a', 'b'], ['c'], ['d', 'e', 'f']]

function allPossibleCases(arr) {
  if (arr.length == 1) {
    return arr[0];
  } else {
    var result = [];
    var allCasesOfRest = allPossibleCases(arr.slice(1));  // recur with the rest of array
    for (var i = 0; i < allCasesOfRest.length; i++) {
      for (var j = 0; j < arr[0].length; j++) {
        result.push(arr[0][j] + allCasesOfRest[i]);
      }
    }
    return result;
  }

}

您也可以使用循环，但这会有些棘手，并且需要实现自己的堆栈模拟。

You can also make it with loops, but it will be a bit tricky and will require implementing your own analogue of stack.

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