const ary1 = [1, 2, 3, 4, 5]; 
const ary2 = [2, 3, 4, 6]; 

• ary1与ary2的并集：Array.from(new Set([ …ary1, …ary2 ]));

• ary1与ary2的交集：ary1.filter(val => ary2.includes(val));

• ary1与ary2的差集: ary1.filter(val => !ary2.includes(val));