Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
336 views
in Technique[技术] by (71.8m points)

algorithm - Finding All Combinations (Cartesian product) of JavaScript array values

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

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'];

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

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

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

EDIT: 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 results = allPossibleCases(allArrays);
 //outputs ["acd", "bcd", "azd", "bzd", "ace", "bce", "aze", "bze", "acf", "bcf", "azf", "bzf"]
Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

This is not permutations, see permutations definitions from Wikipedia.

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;
  }

}

console.log(allPossibleCases(allArrays))

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...