请教一个补集的简单高效算法

Question

<script language=javascript>
var a=new Array('a','b','c');
var b=new Array('a','b','d','e');
var c=new Array();
var temp=","+b.join(",")+",";
var re;
for(var i=0;i<a.length;i++)
{
re=new RegExp(","+a[i]+",")
if(re.test(temp))
{
temp=temp.replace(re,",")
}
}
c=temp.substr(1,temp.length-2).split(",")
alert(c)
</script>

Answer 1

上面那个是补集，再来个交集
<script language=javascript>
var a=new Array('a','b','c');
var b=new Array('a','b','d','e');
var c=new Array();
var temp=","+b.join(",")+",";
var re;
for(var i=0;i<a.length;i++)
{
re=new RegExp(","+a[i]+",")
if(re.test(temp))
{
c.push(a[i])
}
}
alert(c)
</script>

Answer 2

最上面那个还可以在优化下
<script language=javascript>
var b=new Array('a','b','c');
var a=new Array('a','b','d','e');
var c=new Array();
var temp=","+b.join(",")+",";
var re;
for(var i=0;i<a.length;i++)
{
re=new RegExp(","+a[i]+",")
if(!re.test(temp))
{
c.push(a[i])
}

}
alert(c)
</script>

Answer 3

对不起,要的是补集
数组A :{a,b,c}
数组B :{a,b,d,e}
那么我们要的结果是{d,e}
注意A中元素不一定在B中存在