Charles_G

```javascript str.split("").reverse().reduce((prev, cur, index) => (index + 1) % 3 == 0 ? '.' + cur + prev : cur + prev) ```

思路：先将传入的参数转为string,取该字符串的第二位至末尾作为下一次递归函数的参数，判断当前字符串长度是否大于1，大于1则递归调用，等于1就返回这个字符串 ```javascript function myReverse(num){ let str = num.toString() let num2 = parseInt(str.slice(1)) if(str.length >1){ return myReverse(num2) + str[0] }else { return str } } console.log(myReverse(1234)) ```

```javascript function fun1(str) { let obj = {} for (let i = 0; i < str.length; i++) { let key = str[i] if (str[i] == str[i + 1]) { if...

先定义一个判断是否相同的函数，类型不同直接返回false，两边为object则递归再次比较 ```javascript function isOne(left, right) { if (typeof left !== typeof right) { return false } if (typeof left == 'object') { let isTrue = true Object.keys(left).forEach(key => { if...

```javascript function fun1(obj, str = '', result = {}) { Object.keys(obj).forEach((key) => { if (typeof obj[key] == 'object') { fun1(obj[key], str + key + '.', result) } else { str...

Ok, thank you, I've got it.