const intersection = (a, b) => { const s = new Set(b) return [...new Set(a)].filter(x => s.has(x)) }
intersection([1, 2, 3], [4, 3, 2]); // [2, 3]
posted on 2022-09-07 10:09 小馬過河﹎ 阅读(19) 评论(0) 收藏 举报