本文介绍了通过JS的n个列表的交集的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想找到n个列表之间的交集 假设我有一个(正常工作)交叉点(a,b)函数 假设交点( )只需要两个列表作为输入
我正在研究一种算法并试图找出如何解决它,并给出以下信息:
I am working on an algorithm and trying to figure out how to solve it given the following information:
所以问题看起来像这样:
So the problem would look something like this:
var a = {1, 2, 'b'}; var b = {2, 'b', 'b'}; var c = {2, 'b', 'c'}; var d = {'a', 'b', 'c'}; //this is the part that does not work, of course: function intersect_all(d) { //what goes in here??? }注意:我不想为此使用python,因为python有lang中内置的方法,我的应用程序无法使用(或js,就此而言)。我想用上面的信息解决它。
Note: I don't want to use python for this, since python has methods built into the lang that are not available for my app (or js, for that matter). I would like to solve it using the above information.
结果应该类似于
{2, 'b'}jml
推荐答案假设您有一系列列表:
var lists = []; lists[0] = [1, 2, 'b']; lists[1] = [2, 'b', 'b']; lists[2] = [2, 'b', 'c']; lists[3] = ['a', 'b', 'c'];然后你可以使用这个:
// say you call this passing the array of lists as the argument: intersect_all(lists) function intersect_all(lists) { if (lists.length == 0) return []; else if (lists.length == 1) return lists[0]; var partialInt = lists[0]; for (var i = 1; i < lists.length; i++) { partialInt = intersection(partialInt, lists[i]); } return partialInt; }更多推荐
通过JS的n个列表的交集
发布评论