Codewars-Javascript训练手册：字符串（上）

素颜马尾好姑娘i 2022-07-30 14:28 56阅读 0赞

## Recursive reverse string(递归反转字符串) ##

> Description:  
> Your objective is to complete a recursive function reverse() that receives str as String and returns the same string in reverse order

**Rules:**

*  reverse function should be executed only N times. *N = length of the input string*
 *  helper functions are not allowed
 *  changing the signature of the function is not allowed

**stackoverflow解决方案：**

function reverse (str) {
        if (str === "") {
            return "";
        } else {
            return reverse(str.substr(1)) + str.charAt(0);
        }
    
    }

可改写为：`return str=="" ? "":reverse(str.substr(1)) + str.charAt(0)`,`substr()` 方法可在字符串中抽取从 start 下标开始的指定数目的字符。语法：

> stringObject.substr(start,length)

ECMAscript 没有对该方法进行标准化，因此反对使用它。 `slice()` 可以代替`substr()`.`charAt()` 方法可返回指定位置的字符,注释：字符串中第一个字符的下标是 0。如果参数 `index` 不在 0 与 `string.length` 之间，该方法将返回一个空字符串。可以使用`str[0]`方法代理

**Codewars解决方案：**

function reverse(str) {
      return str.length == 1 ? str : reverse(str.substr(1)) + str[0];
    }

`slice()` 方法可提取字符串的某个部分，并以新的字符串返回被提取的部分。语法：

> stringObject.slice(start,end)

## 字符串去重 ##

> Description:  
> Implement the function unique\_in\_order which takes as argument a sequence and returns a list of items without any elements with the same value next to each other and preserving the original order of elements.

**Solution:**

## Dubstep ##

> **Description:**  
> a song with words “I AM X” can transform into a dubstep remix as “WUBWUBIWUBAMWUBWUBX” and cannot transform into “WUBWUBIAMWUBX”.  
> **Output**  
> Return the words of the initial song that Polycarpus used to make a dubsteb remix. Separate the words with a space.

**Solution:**

**Codewars上的解决方案：**

//方案一
    function songDecoder(song){ 
      return song.replace(/(WUB)+/g," ").trim()
    }
    //方案二
    function songDecoder(song){ 
      return song.split("WUB").filter(function(v){ return v!==""}).join(" ");
    }