Javascriptツリートラバーサルアルゴリズム

Question

深みのある方法でツリー構造を横断する助けが必要です。私はそれを正しく行うためのアルゴリズムを考え出すことができません。

私の入力はこれです：出力形式をとるべき

[ 
    ["A", "B", "C"], 
    ["1", "2"], 
    ["a", "b", "c", "d"] 
] 

：

[ 
    "A/1/a", "A/1/b", "A/1/c", "A/1/d", 
    "A/2/a", "A/2/b", "A/2/c", "A/2/d", 
    "B/1/a", "B/1/b", "B/1/c", "B/1/d", 
    "B/2/a", "B/2/b", "B/2/c", "B/2/d", 
    "C/1/a", "C/1/b", "C/1/c", "C/1/d", 
    "C/2/a", "C/2/b", "C/2/c", "C/2/d" 
] 

Answer 1

これは、ジョブを実行する必要があります。

function traverse(arr) { 
 
    var first = arr[0]; 
 
    var ret = []; 
 
    if (arr.length == 1) { 
 
    for (var i = 0; i < first.length; i++) { 
 
     ret.push(first[i]); 
 
    } 
 
    } else { 
 
    for (var i = 0; i < first.length; i++) { 
 
     var inn = traverse(arr.slice(1)); 
 
     for (var j = 0; j < inn.length; j++) { 
 
     ret.push(first[i] + '/' + inn[j]); 
 
     } 
 
    } 
 
    } 
 
    return ret; 
 
} 
 

 
var inp = [ 
 
    ["A", "B", "C"], 
 
    ["1", "2"], 
 
    ["a", "b", "c", "d"] 
 
]; 
 

 
var out = traverse(inp); 
 

 
console.log(out);

Answer 2

あなたが探しているものは、リストのリストの直積である、それはaskedてきました前。その質問のための受け入れ答えから借りて、あなたはJavascript 1.7でこれを行うことができます。

function product() { 
    return Array.prototype.reduce.call(arguments, function(as, bs) { 
     return [a.concat(b) for each (a in as) for each (b in bs)]; 
    }, [[]]); 
}; 

function convert(lst) { 
    var solution = []; 
    for (var i = 0; i < lst.length; i++) { 
    solution.push(lst[i][0] + "/" + lst[i][1] + "/" + lst[i][2]); 
    } 
    return solution; 
}; 

convert(product(["A", "B", "C"], ["1", "2"], ["a", "b", "c", "d"])); 

> ["A/1/a", "A/1/b", "A/1/c", "A/1/d", 
    "A/2/a", "A/2/b", "A/2/c", "A/2/d", 
    "B/1/a", "B/1/b", "B/1/c", "B/1/d", 
    "B/2/a", "B/2/b", "B/2/c", "B/2/d", 
    "C/1/a", "C/1/b", "C/1/c", "C/1/d", 
    "C/2/a", "C/2/b", "C/2/c", "C/2/d"]