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Asked: 2026-06-14T00:30:16+00:00

My problem here is that i want a Character remove in some parts of

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My problem here is that i want a Character remove in some parts of a String but I do not know how to restrict the removing.

Example:

A computer is a general purpose device that can be\n
programmed to carry out a finite set of\n 
millions to billions of times more capable.\n
\n
In this era mechanical analog computers were used\n
for military applications.\n
1.1 Limited-function early computers\n
1.2 First general-purpose computers\n
1.3 Stored-program architecture\n
1.4 Semiconductors and\n

this here example is the content of my string, what i want to happen is to remove the \n of lines 1 and 2 above but not to remove the \n in line 5 onwards. How do i remove the \n without removing the other \n?. My Goal here is to make the string a paragraph without \n after line. like the example the first 3 lines can be a paragraph and the next lines are in bullet form(example). what i am saying is that I do not want to remove \n in bulleted characters.

The real contents of the string is dynamic.

I have tried using String.replaceAll("\n", " ") well clearly that would not work it will remove all the \n i have thought of using Regex to determine what is Alphanumberic but it would remove some letters after \n

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1 Answer

  1. Editorial Team

    Try using this regex: –

    str = str.replaceAll("(.+)(?<!\\.)\n(?!\\d)", "$1 ");
System.out.println(str);

    This will replace your \n if it is not preceded by a dot - termination of a paragraph, and it is not followed by a digit, for when it is followed by a bulleted point. (like, your \n in first bullet point is followed by a 1.2. So, it will not be replaced.).

    • (.+) at the start, ensures that you are not replacing a blank line.

    This will work for the string you have shown.

    Explanation: –

    • (.+) -> A capture group, capturing anything, occurring at least once.

    • (?<!\\.) -> This is called negative-look-behind. It matches the string following it, only if that string is not preceded by a dot(.) given in the negative-look-behind pattern.
      For e.g.: – You don’t need to replace \n after the line: – millions to billions of times more capable.\n.

    • (?!\\d) -> This is called negative -look-ahead. It matches string behind it, only if that string is not followed by a digit (\\d) given in the negative-look-ahead pattern.
      For e.g.: – In your bulleted points, computers\n is followed by 1.2. where 1 is a digit. So, you don’t want to replace that \n.

    • Now, $1 and $2 represent the groups captured in the pattern match. Since you just want to replace "\n". So, we took the remaining pattern match as it is, while replacing "\n" with a space.

    So, $1 is representation for 1st group – (.+)

    Note, look-ahead and look-behind regexes are non-capturing groups.

    For More Details, follow these links: –

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