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How to remove white-space at the front of a string using string split in java?

I was solving a problem on string.split(), but I couldn't find the solution to remove white-space at the start. Actually the problem is in the regular expression. It needs to be changed. I tried to solve by changing the expression several times but didn't work out. What will be the regular expression? Here is the code given below:
String s = " YES leading spaces are valid, problemsetters are evillllll";
String delims = "[\\s._,?!'#\\t]+";
String[] words = s.split(delims);
System.out.println(words.length - 1);
for(String w:words) {
System.out.println(w);
}
System.out.println(Arrays.toString(words));
output:
8
YES
leading
spaces
are
valid
problemsetters
are
evillllll
[, YES, leading, spaces, are, valid, problemsetters, are, evillllll]

You can use String class trim() method to remove the space before using split

Java Split regex

Given a string S, find the number of words in that string. For this problem a word is defined by a string of one or more English letters.
Note: Space or any of the special characters like ![,?.\_'#+] will act as a delimiter.
Input Format: The string will only contain lower case English letters, upper case English letters, spaces, and these special characters: ![,?._'#+].
Output Format: On the first line, print the number of words in the string. The words don't need to be unique. Then, print each word in a separate line.
My code:
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
String regex = "( |!|[|,|?|.|_|'|#|+|]|\\\\)+";
String[] arr = str.split(regex);
System.out.println(arr.length);
for(int i = 0; i < arr.length; i++)
System.out.println(arr[i]);
When I submit the code, it works for just over half of the test cases. I do not know what the test cases are. I'm asking for help with the Murphy's law. What are the situations where the regex I implemented won't work?

You don't escape some special characters in your regex. Let's start with []. Since you don't escape them, the part [|,|?|.|_|'|#|+|] is treated like a set of characters |,?._'#+. This means that your regex doesn't split on [ and ].
For example x..]y+[z is split to x, ]y and [z.
You can fix that by escaping those characters. That will force you to escape more of them and you end up with a proper definition:
String regex = "( |!|\\[|,|\\?|\\.|_|'|#|\\+|\\])+";
Note that instead of defining alternatives, you could use a set which will make your regex easier to read:
String regex = "[!\\[,?._'#+\\].]+";
In this case you only need to escape [ and ].
UPDATE:
There's also a problem with leading special character (like in your example ".Hi?there[broski.]#####"). You need to split on it but it produces an empty string in the results. I don't think there's a way to use split function without producing it but you can mitigate it by removing the first group before splitting using the same regex:
String[] arr = str.replaceFirst(regex, "").split(regex);

How to use split() to remove all delimiters from a sentence in Java?

String text = "Good morning. Have a good class. " +
"Have a good visit. Have fun!";
String[] words = text.split("[ \n\t\r.,;:!?(){");
This split method is provided in text book, meant to remove all the delimiters in the sentence as well as white space character but clearly it is not working and throws a regex exception to my disappointment....I am wondering what could we do here to make it work? The requirement is after the split method, everything in the `String[] words are either just English words without any delimiters attaching to it or whitespace character! Thanks a lot!

You are missing closing ] in your character class:
String[] words = text.split("[ \n\t\r.,;:!?(){]");
btw you can just do (and it is better option):
String[] words = text.split("\\W+");
to split on any non-word character.

String.split() is NOT for removing characters. It is used to divide the String into smaller substrings.
Example:
String s = "This is a string!";
String[] tokens = s.split(" ");
Split will have used the String " " (one space character) as a delimiter to, well, split the string. As a result, the array tokens will look something like
{"This", "is", "a", "string!"}
If you want to remove characters, try taking a look at String.replaceAll()

Java String delete tokens contains numbers

I have a string like this and I would like to eliminate all the tokens that contain a number:
String[] s="In the 1980s".split(" ");
Is there a way to remove the tokens that contain numbers - in this case 1980s, but also, for example 784th or s787?

Use a \w*\d\w* regex matcher for that. It will match all words with at least one digit in them. Although I generally despise regexes, they are particularily well suited for your problem.
String[] s = input.replaceAll("\\w*\\d\\w* *", "").split(" +");
See Java lib docs for Pattern/Matcher (RegEx) for more reference how to work with regexes in general.
Test code:
http://ideone.com/LrHDsT

Remove the unwanted words first, then split:
String[] s = str.replaceAll("\\w*\\d\\w*", "").trim().split(" +");
Some test code:
String str = "666 In the 1980s 784th s787 foo BAR";
String[] s = str.replaceAll("\\w*\\d\\w*", "").trim().split(" +");
System.out.println(Arrays.toString(s));
Output:
[In, the, foo, BAR]

You could Regex as suggested by #vaxquis or alternately after splitting the string based on the delimiter
You could Parse the token strings and check if the token has number among them using NumberUtils.isNumber and remove those tokens.

split doesn't seem to be what you are looking for. Even if you remove words which contain digit like in case of
"1foo f2oo bar whatever baz2"
you will end up with
" bar whatever "
and if you split on spaces now you will end up with ["", "bar", "whatever"].
To solve this problem you may want also to remove spaces after word you removed so now
"1foo f2oo bar whatever baz2"
would become
"bar whatever "
so it can be split correctly (space at the end is not the problem since split by default removes trailing empty strings in result array).
But instead of doing two iterations (removing words and splitting on string) you can achieve same thing with only one iteration. All you need to do is use opposite approach:instead of focusing on removing wrong elements, lets try to find correct ones.
Correct tokens seem to be words which contains any non-space characters but not digits. You can regex representing such words with this regex \b[\S&&\D]\b where:
\b represents word boundaries,
\S any non whitespace character
\D any non digit character
[\S&&\D] intersection of non-whitespaces and non-digits, in other words non whitespaces which are also non-ditigts
Demo:
String input = "1foo f2oo bar whatever baz2";
Pattern p = Pattern.compile("\\b[\\S&&\\D]+\\b");
Matcher m = p.matcher(input);
while(m.find())
System.out.println(m.group());
Output:
bar
whatever
BTW to avoid potential problems with potential empty element at start of results you can use Scanner which doesn't return empty element if delimiter is found at start of string. So we can simply set delimiter as series of spaces or words which contains digit. So your code can also look like
Scanner sc = new Scanner(input);
sc.useDelimiter("(\\s|\\w*\\d\\w*)+");
while (sc.hasNext())
System.out.println(sc.next());
sc.close();

Java (Regex) - Get all words in a sentence

I need to split a java string into an array of words. Let's say the string is:
"Hi!! I need to split this string, into a serie's of words?!"
At the moment I'm tried using this String[] strs = str.split("(?!\\w)") however it keeps symbols such as ! in the array and it also keeps strings like "Hi!" in the array as well. The string I am splitting will always be lowercase. What I would like is for an array to be produced that looks like:
{"hi", "i", "need", "to", "split", "this", "string", "into", "a", "serie's", "of", "words"} - Note the apostrophe is kept.
How could I change my regex to not include the symbols in the array?
Apologies, I would define a word as a sequence of alphanumeric characters only but with the ' character inclusive if it is in the above context such as "it's", not if it is used to a quote a word such as "'its'". Also, in this context "hi," or "hi-person" are not words but "hi" and "person" are. I hope that clarifies the question.

You can remove all ?! symbols and then split into words
str = str.replaceAll("[!?,]", "");
String[] words = str.split("\\s+");
Result:
Hi, I, need, to, split, this, string, into, a, serie's, of, words

Should work for what you want.
String line = "Hi!! I need to split this string, into a serie's of words?! but not '' or ''' word";
String regex = "([^a-zA-Z']+)'*\\1*";
String[] split = line.split(regex);
System.out.println(Arrays.asList(split));
Gives
[Hi, I, need, to, split, this, string, into, a, serie's, of, words, but, not, or, word]

If you define a word as a sequence of non-whitespace characters (whitespace character as defined by \s), then you can split along space characters:
str.split("\\s+")
Note that ";.';.##$>?>#4", "very,bad,punctuation", and "'goodbye'" are words under the definition above.
Then the other approach is to define a word as a sequence of characters from a set of allowed characters. If you want to allow a-z, A-Z, and ' as part of a word, you can split along everything else:
str.split("[^a-zA-Z']+")
This will still allow "''''''" to be defined as a word, though.

So what you want is to split on anything that is not a wordcharacter [a-zA-Z] and is not a '
This regex will do that "[^a-zA-Z']\s"
There will be a problem if the string contains a quote that is quoted in '
I usually use this page for testing my regex'
http://www.regexplanet.com/advanced/java/index.html

I would use str.split("[\\s,?!]+"). You can add whatever character you want to split with inside the brackets [].

You could filter out the characters you deem as "non-word" characters:
String[] strs = str.split("[,!? ]+");

myString.replaceAll("[^a-zA-Z'\\s]","").toLowerCase().split("\\s+");
replaceAll("[^a-zA-Z'\\s]","") method replaces all the characters which are not a-z or A-Z or ' or a whitespace with nothing ("") and then toLowerCase method make all the chars returned from replaceAll method lower case. Finally we are splitting the string in terms of whitespace char. more readable one;
myString = myString.replaceAll("[^a-zA-Z'\\s]","");
myString = myString.toLowerCase();
String[] strArr = myString.split("\\s+");