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生成所有可能的字符串组合

[英]Generating all possible combinations of strings

I am trying to generate all possible combinations of a string. 我正在尝试生成字符串的所有可能组合。

eg for the list below: a1q5z!H9, b1q5z!H9, c1q5z!H9, d1q5z!H9, a2q5z!H9 ... etc 例如,对于下面的列表:a1q5z!H9,b1q5z!H9,c1q5z!H9,d1q5z!H9,a2q5z!H9 ......等

Rather than make lots of nested loops, I thought I would try something clever with MODULO ... but hit a wall. 而不是制作大量嵌套循环,我想我会尝试用MODULO聪明的东西......但是碰壁了。

This is the Javascript I have come up with - any pointers to how I might go on? 这是我想出的Javascript - 我可能会怎么做的任何指示?

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. 这只是输出列表中的前20个,但重复...我几乎拥有它但不完全。 Any help or pointers appreciated. 任何帮助或指针赞赏。

Quite easy to write a recursive function demo . 编写递归函数演示非常容易。

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 或者更迫切的方法

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;
}

This is how I did it: 我就是这样做的:

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. blah是字符矩阵。

Ugly, granted, but the shorter answer by Yury hurts my head. 丑陋,被授予,但Yury的简短回答伤害了我的头脑。

function combinations(str) {

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

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

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