what is the fastest way to replace all appearances of ... in strings by space? Java

Question

I have problem in my program. In my program a bottleneck is replacing and spliting String. I need to get words to tab from String.

For example:

I have String: "This is Ala. Does Ala have a cat? Money-making cat."

I need to get a String tab[] and results

tab[0]="This"<br>
tab[1]="is"<br>
tab[2]="Ala"    not "Ala."<br>
tab[3]="Does"<br>
....<br>
tab[7]="cat" not "cat?"<br>
tab[8]="Money"   not "Money-making"<br>
tab[9]="making"<br>
tab[10]="cat" not "cat."  <br>

The words cant have signs like ",./;!:?- etc. They can have only english letters.

Actually im doing this like that

s = s.replace(",", " ").replace("!", " ")....   ;

String [] tab = s.split("\\s+");

But this way is really slow. How can i do that faster? In Java Language.

Answer 1

正则表达式是您的朋友。

s = s.replaceAll("[^a-zA-Z]"," ");

Answer 2

You can split at one or more non-word characters:

String[] parts = str.split("\\W+");

Note: Non-word characters mean anything other than _ , letters and digits. If you only want lettes than you would have to go with @Bailey S 's answer.

Answer 3

您可以使用replaceAll。例如s.replaceAll（“ [？。，]”，“”）