拆分正则表达式以提取连续字符的字符串 [英] Split regex to extract Strings of contiguous characters

问题描述

是否有适用于 String.split（） 将字符串分成连续的字符 - 即分割下一个字符与前一个字符不同的位置？

Is there a regex that would work with String.split() to break a String into contiguous characters - ie split where the next character is different to the previous character?

以下是测试用例：

    String regex = "your answer here";
    String[] parts = "aaabbcddeee".split(regex);
    System.out.println(Arrays.toString(parts));

预期产出：

[aaa, bb, c, dd, eee]

虽然测试用例只有字母作为输入，这只是为了清晰起见;输入字符可能是任何字符。

Although the test case has letters only as input, this is for clarity only; input characters may be any character.

请不要提供解决方法循环或其他技术。

Please do not provide "work-arounds" involving loops or other techniques.

问题是为代码找到正确的正则表达式，如上所示 - 即仅使用 split（），没有其他方法调用。这不是一个关于找到能够完成工作的代码的问题。

The question is to find the right regex for the code as shown above - ie only using split() and no other methods calls. It is not a question about finding code that will "do the job".

推荐答案

完全可以编写正则表达式分步一步：

It is totally possible to write the regex for splitting in one step:

"(?<=(.))(?!\\1)"

由于你想要在每组相同的字符之间进行拆分，我们只需要寻找2之间的边界。组。我通过使用正面的后视来获取前一个字符，并使用负前瞻和后引用来检查下一个字符是不是相同的字符。

Since you want to split between every group of same characters, we just need to look for the boundary between 2 groups. I achieve this by using a positive look-behind just to grab the previous character, and use a negative look-ahead and back-reference to check that the next character is not the same character.

正如您所看到的，正则表达式是零宽度（只有2个查看断言）。正则表达式不会消耗任何字符。

As you can see, the regex is zero-width (only 2 look around assertions). No character is consumed by the regex.

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