Regex expressions in Java, \\s vs. \\s+

What's the difference between the following two expressions?

x = x.replaceAll("\\s", "");
x = x.replaceAll("\\s+", "");

Solution 1:

The first one matches a single whitespace, whereas the second one matches one or many whitespaces. They're the so-called regular expression quantifiers, and they perform matches like this (taken from the documentation):

Greedy quantifiers
X?  X, once or not at all
X*  X, zero or more times
X+  X, one or more times
X{n}    X, exactly n times
X{n,}   X, at least n times
X{n,m}  X, at least n but not more than m times

Reluctant quantifiers
X?? X, once or not at all
X*? X, zero or more times
X+? X, one or more times
X{n}?   X, exactly n times
X{n,}?  X, at least n times
X{n,m}? X, at least n but not more than m times

Possessive quantifiers
X?+ X, once or not at all
X*+ X, zero or more times
X++ X, one or more times
X{n}+   X, exactly n times
X{n,}+  X, at least n times
X{n,m}+ X, at least n but not more than m times

Solution 2:

Those two replaceAll calls will always produce the same result, regardless of what x is. However, it is important to note that the two regular expressions are not the same:

• \\s - matches single whitespace character
• \\s+ - matches sequence of one or more whitespace characters.

In this case, it makes no difference, since you are replacing everything with an empty string (although it would be better to use \\s+ from an efficiency point of view). If you were replacing with a non-empty string, the two would behave differently.