Question

我有一个25个字符的代码，我知道一些字符，但我遗漏了剩下的字符。到目前为止，我有以下代码来帮助解决丢失的代码选项：

function combinations(w) {
var x = new Array(
         "B","C","D","F","G","H","J","K",
         "M","P","Q","R","T","V",
         "W","X","Y","Z",2,3,4,6,7,8,9
    ),
    n = w.match(/\!/g).length,
    x_n = new Array(),
    r = new Array(),
    c = null;

for (var i = n; i > 0; i--) {
    x_n.push(x);
}

c = x_n.reduce(function(a, b) {
    var c = [];
    a.forEach(function(a) {
        b.forEach(function(b) {
            c.push(a.concat([b]));
        });
    });
    return c;
}, [[]]);

for (var i = 0, j = 0; i < c.length; i++, j = 0) {
    r.push(w.replace(/\!/g, function(s, k) {
        return c[i][j++];
    }));
}

return r;}

console.log(combinations("6X3#!-#!KHY-BgVDT-W!WDV-H6HJR"))

对于25个字符的代码＆＃34; 7X3 !! - ＃！KHY-BKVDT-W！QDV-H6HFR＆＃34;这段代码可以很好地列出所有空格的所有可能结果，使用＆＃34;！＆＃34;。

我想进一步改进我的结果。我试图找出如何使用＆＃34;！＆＃34;从一组可能性中进行选择，并使用＆＃34;＃＆＃34;从不同的可能性中选择，并最终得到所有可能结果的列表。

第二组可能性如下：

var p = new Array(
        "B","D",
        "P","R",
         2,8)

Javascript for Variations with Repetition缺少具有多个选项数组的字符串字符

0 个答案: