Codewars 刷題第二天，仍為字串問題：

第二期題目：

Polycarpus works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.

Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Polycarpus inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.

For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".

Recently, Jonny has heard Polycarpus's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Polycarpus remixed. Help Jonny restore the original song.

Input

The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters

Output

Return the words of the initial song that Polycarpus used to make a dubsteb remix. Separate the words with a space.

Examples

songDecoder("WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB")  // =>  WE ARE THE CHAMPIONS MY FRIEND

 

題目要求：

            對長度不超過200的字串進行某一子串的處理，意味著將字串中的WUB字串去掉，恢復字串原有的形式；

解析：    

1. 1.  子串意味著連續的三個字元，可以遍歷，判斷子串是否出現，並處理；

   2.   也可以直接用串匹配的方法進行處理！

答案(一）：（已AC)

public static String SongDecoder (String song){

    //     Your code is here...

    return song.replaceAll("(WUB)+", "  ").trim();

    }

}

答案（二）：已AC

public static String SongDecoder (String song){

if (song.length() > 200 || song.isEmpty()) {

return "";

}

char[] songCharArr = song.toCharArray();

StringBuilder strBuilder = new StringBuilder();

boolean buildingWord = false; 

int i = 0;

while (i < song.length())  {

if ((i < song.length()-2) && songCharArr[i] == 'W' && songCharArr[i+1] == 'U' && songCharArr[i+2] == 'B'){

if (buildingWord){

strBuilder.append(" ");

buildingWord = false;

}

i += 3;

} else {

strBuilder.append(songCharArr[i]);

buildingWord = true;

i += 1;

}

}

return strBuilder.toString().trim();

}

答案一：正則表示式使用直接替換掉（WUB)+ 在trim()掉首尾的空格；

答案二：通常的思路分析串的長度，以及WUB所在的位置是否為串頭，串尾，去掉之後是否為可輸出的內容（是否加空格）；