通过JS的n个列表的交集 [英] intersection of n lists via JS

查看:160
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:


  1. 我想找到n个列表之间的交集

  2. 假设我有一个(正常工作)交叉点(a,b)函数

  3. 假设交点( )只需要两个列表作为输入

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;
}

这篇关于通过JS的n个列表的交集的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆