求star

①递归处理:milky_way:#

Array.prototype.flatten = function () {
    let res = []
    let len = this.length
    for(let i=0;i<len;i++){
        if(Array.isArray(this[i])){
            res = res.concat(this[i].flatten())
        }else {
            res.push(this[i])
        }
    }
    return res
}

let arr=[1,2,3,[4,5,'hello',['world',9,666]]]
console.log(arr.flatten())	// [1, 2, 3, 4, 5, 'hello', 'world', 9, 666]

②数组的reduce方法:milky_way:#

function flatten(arr) {
	return arr.reduce((prev,item) => {
        return prev.concat(Array.isArray(item) ? flatten(item) : item)
    },[])
}
let arr=[1,2,3,[4,5,'hello',['world',9,666]]]
console.log(flatten(arr))	// [1, 2, 3, 4, 5, 'hello', 'world', 9, 666]

③利用数组的join和split方法:milky_way:#

let arr = [4, 1, 2, 3, 6, [7, 8, [3, 9, 10, [4, 6, 11]]]]
let res = arr.join(',').split(',').map(Number)
console.log(res)	// [4, 1, 2, 3, 6, 7, 8, 3, 9, 10, 4, 6, 11]

④利用正则方法:milky_way:#

let arr = [4, 1, 2, 3, 6, [7, 8, [3, 9, 10, [4, 6, 11]]]]
let res = arr.join(',').replace(/\[|\]/g,"").split(',').map(Number)
console.log(res)	// [4, 1, 2, 3, 6, 7, 8, 3, 9, 10, 4, 6, 11]

⑤ES6新方法flat():milky_way:#

TIP

​ 实现数组的”拉平”(扁平化),该方法返回一个新数组，对原数据没有影响。

​ 具有一个可选参数depth,用来指定深度,默认为1

​ 传入’Infinity’可以拉平任意深度的数组

let arr =  [4, 1, 2, 3, 6, [7, 8, [3, 9, 10, [4, 6, 11]]]]
console.log(arr.flat())	// [4, 1, 2, 3, 6, 7, 8, [3, 9, 10, [4, 6, 11]]]
console.log(arr.flat(2)) // [4, 1, 2, 3, 6, 7, 8, 3, 9, 10, [4, 6, 11]]
console.log(arr.flat(Infinity)) // [4, 1, 2, 3, 6, 7, 8, 3, 9, 10, 4, 6, 11]

⑥利用...扩展运算符循环:milky_way:#

function flatten(arr) {
    while (arr.some(item => Array.isArray(item))) {
        arr = [].concat(...arr);
    }
    return arr;
}
console.log(flatten([4, 1, 2, 3, 6, [7, 8, [3, 9, 10, [4, 6, 11]]]])) // [4, 1, 2, 3, 6, 7, 8, 3, 9, 10, 4, 6, 11]