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I am trying to generate all possible binations of a string.

e.g. for the list below: a1q5z!H9, b1q5z!H9, c1q5z!H9, d1q5z!H9, a2q5z!H9 ... etc

Rather than make lots of nested loops, I thought I would try something clever with MODULO ... but hit a wall.

This is the Javascript I have e up with - any pointers to how I might go on?

var c = [
  ['a', 'b', 'c', 'd'],
  ['1', '2', '3', '4'],
  ['q', 'w', 'e', 'r'],
  ['5', '6', '7', '8'],
  ['z', 'x', 'c', 'v'],
  ['!', '"', '£', '$'],
  ['H', 'J', 'K', 'L'],
  ['9', '8', '7', '6'],
];

var o = document.getElementById('output');
var pw = "";
var chars = c.length;

for( var i = 0; i <20; i++)
{
  pw = ""
  for(var j = 0; j < chars; j++ )
    {
      pw += c[j][i%4];
    }
  op(pw);
}

function op(s)
{
  o.innerHTML = o.innerHTML + "<br>" + s;
}

This just outputs the first 20 in the list, but repeats ... I nearly have it but not quite. Any help or pointers appreciated.

I am trying to generate all possible binations of a string.

e.g. for the list below: a1q5z!H9, b1q5z!H9, c1q5z!H9, d1q5z!H9, a2q5z!H9 ... etc

Rather than make lots of nested loops, I thought I would try something clever with MODULO ... but hit a wall.

This is the Javascript I have e up with - any pointers to how I might go on?

var c = [
  ['a', 'b', 'c', 'd'],
  ['1', '2', '3', '4'],
  ['q', 'w', 'e', 'r'],
  ['5', '6', '7', '8'],
  ['z', 'x', 'c', 'v'],
  ['!', '"', '£', '$'],
  ['H', 'J', 'K', 'L'],
  ['9', '8', '7', '6'],
];

var o = document.getElementById('output');
var pw = "";
var chars = c.length;

for( var i = 0; i <20; i++)
{
  pw = ""
  for(var j = 0; j < chars; j++ )
    {
      pw += c[j][i%4];
    }
  op(pw);
}

function op(s)
{
  o.innerHTML = o.innerHTML + "<br>" + s;
}

This just outputs the first 20 in the list, but repeats ... I nearly have it but not quite. Any help or pointers appreciated.

• javascript
• cartesian

Share Improve this question Follow edited Jun 24, 2014 at 13:58 Ruskin asked Jun 24, 2014 at 13:29 RuskinRuskin 6,17144 gold badges4949 silver badges6565 bronze badges 6

• Why is c structured like that? – Cerbrus Commented Jun 24, 2014 at 13:30
• and I know string concatenation is not the most efficient, but this is quick and dirty code - not optimised. – Ruskin Commented Jun 24, 2014 at 13:30
• Quick and dirty array of chars - first thing that came to mind. Data is meaninless - just random chars – Ruskin Commented Jun 24, 2014 at 13:33
• The length of the string has an importance, or "ab2" is a valid result too? – Nicolas Henrard Commented Jun 24, 2014 at 13:38
• Yea, but why is it an array of arrays of chars? Why the nested array? – Cerbrus Commented Jun 24, 2014 at 13:41

 |  Show 1 more ment

3 Answers 3

Sorted by: Reset to default 6

Quite easy to write a recursive function demo.

function permutate(abc, memo) {
    var options;
    memo = memo || abc.shift().slice(0);

    if(abc.length) {
        options = abc.shift();

        return permutate(abc, memo.reduce(function(all, item){
            return all.concat(options.map(function(option){
                return item + option;
            }))
        }, []));       
    }

    return memo;
};

console.log(permutate(c).length); //65536 items

Or more imperative approach

function permutate2(abc) {
    var options, i, len, tmp, j, optionsLen, 
        memo = abc.pop().slice(0); //copy first the last array

    while(options = abc.pop()) { //replace recursion
        tmp = [];
        optionsLen = options.length;
        for(i = 0, len = memo.length; i < len; i++) { //for every element in memo
            for(j = 0; j < optionsLen; j++) { //do cartesian product with options
                tmp.push(options[j] + memo[i]);    
            }
        }
        memo = tmp;
    }

    return memo;
}

function binations(str) {

debugger;
var arr = [];   
for (var i = 0; i < str.length; i++) 
{
    // create an empty string
    var b = "";
    // loop for substring
    for (var j = i; j < str.length; j++) 
        {
        b+=str[j];
        arr.push(b);
        }
}
  return arr;

}
console.log(binations('output'));

This is how I did it:

string password;
bool done = false;

while (!done) {
    password = string.Empty;

    for(int a = 0; a < blah[0].Length; a++) {
        for(int b = 0; b < blah[1].Length; b++) {
            for (int c = 0; c < blah[2].Length; c++) {
                for (int d= 0; d < blah[3].Length; d++) {
                    for (int e = 0; e < blah[4].Length; e++) {
                        for (int f = 0; f < blah[5].Length; f++) {
                            for (int g = 0; g < blah[6].Length; g++) {
                                for (int h = 0; h < blah[7].Length; h++) {
                                    password = string.Format(
                                        "{0}{1}{2}{3}{4}{5}{6}{7}", 
                                        blah[0][a], 
                                        blah[1][b], 
                                        blah[2][c], 
                                        blah[3][d], 
                                        blah[4][e], 
                                        blah[5][f], 
                                        blah[6][g], 
                                        blah[7][h]);

                                    Console.Out.WriteLine(password);
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}

Where blah is the character matrix.

Ugly, granted, but the shorter answer by Yury hurts my head.

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