es6的Set解构

let arr = [1,3,6,3,1,9]

let arr1 = new Set(arr)

es6 set解构轻松去重

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let a = new Set([1, 2, 3]);

let b = new Set([4, 3, 2]);

// 并集let union = new Set([...a, ...b]);// Set {1, 2, 3, 4}

// 交集let intersect = new Set([...a].filter(x => b.has(x)));// set {2, 3}

// 差集let difference = new Set([...a].filter(x => !b.has(x)));