Reduce三种常用的场景

//将二维数组转换成一维数组(将多维转化成一维)

  //将二维数组转换成一维数组
    let arr = [[0, 1], [2, 3], [4, 5]]
    let newArr = arr.reduce((pre, cur) => {
        return pre.concat(cur)
    }, [])
    console.log(newArr);
    //将多维转化成一维
    let arr1 = [[0, 1], 2, [3, 4, [5, 6]], 7]

    const newArr2 = function (arr) {
        return arr.reduce((pre, cur) => pre.concat(Array.isArray(cur) ? newArr2(cur) : cur), [])
    }
    console.log(newArr2(arr1));

 //判断字符串字符出现的次数

  let str = 'aaaabbbbbbbbcccccccccccccc'
    function fn(str) {
        let arr = str.split('')
        return arr.reduce((pre, cur) => {
            if (cur in pre) {
                pre[cur]++
            } else {
                pre[cur] = 1
            }
            return pre
        }, {})
    }
    console.log(fn(str));

    //以上我们统计了 字符串中所有的字母极其出现的次数
    //记下来 我们求出出现最多的
    let obj = fn(str)
    let maxStr = null
    let Max = 0
    for (var cur in obj) {
        if (Max < obj[cur]) {
            Max = obj[cur]
            maxStr = cur
        }
    }
    console.log(`出现次数最多的是${maxStr},出现的次数是${Max}次`);

//数组去重

        

 let arr = [1, 1, 2, 5, 6, 3, 3, 3, 3, 4]


    let newArr = arr.reduce((pre, cur) => {
        // console.log(cur);
        if (!pre.includes(cur)) {
            return pre.concat(cur)
        } else {
            return pre
        }
    }, [])


    let newArr2 = arr.reduce((pre, cur) => {
        // console.log(pre);
        if (pre.indexOf(cur) == -1) {
            return pre.concat(cur)
        } else {
            return pre
        }
    }, [])

    console.log(newArr);
    console.log(newArr2);