[英]Having stack overflow problems in a tree traversal algorithm for an interpreter in JavaScript?
[英]Javascript Tree Traversal Algorithm
我需要以深度優先的方式遍歷樹結構的幫助。 我無法想出一個算法來正確地做到這一點。
我的意見是這樣的:
[
["A", "B", "C"],
["1", "2"],
["a", "b", "c", "d"]
]
輸出應采取以下形式:
[
"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"
]
這應該做的工作:
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);
您正在尋找的是列表清單中的笛卡爾積,之前已經被問過。 借用該問題的已接受答案,您可以在Javascript 1.7中執行此操作:
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"]
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