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In a recent question (Remove whitespaces in a stringRemove whitespaces in a string) my solution included using a simple regex replacement (\\s+) to match all instances of double spaces (or more) and reduce them to a single space.

While this works great its been bugging me that the code might not be a little more efficient to use (\\s\\s+) as this would reduce the number of matches requiring replacement.

Short Question

Is it more efficient to use a more complex regex match when doing a replaceAll()

In a recent question (Remove whitespaces in a string) my solution included using a simple regex replacement (\\s+) to match all instances of double spaces (or more) and reduce them to a single space.

While this works great its been bugging me that the code might not be a little more efficient to use (\\s\\s+) as this would reduce the number of matches requiring replacement.

Short Question

Is it more efficient to use a more complex regex match when doing a replaceAll()

In a recent question (Remove whitespaces in a string) my solution included using a simple regex replacement (\\s+) to match all instances of double spaces (or more) and reduce them to a single space.

While this works great its been bugging me that the code might not be a little more efficient to use (\\s\\s+) as this would reduce the number of matches requiring replacement.

Short Question

Is it more efficient to use a more complex regex match when doing a replaceAll()

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Jon Hazan
Jon Hazan
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Regex replacement efficiency

In a recent question (Remove whitespaces in a string) my solution included using a simple regex replacement (\\s+) to match all instances of double spaces (or more) and reduce them to a single space.

While this works great its been bugging me that the code might not be a little more efficient to use (\\s\\s+) as this would reduce the number of matches requiring replacement.

Short Question

Is it more efficient to use a more complex regex match when doing a replaceAll()