codingBat repeatEnd使用正则表达式 [英] codingBat repeatEnd using regex

问题描述

我正在尝试尽可能多地了解正则表达式，所以我想出了这个基于正则表达式的解决方案 codingbat.com repeatEnd ：

I'm trying to understand regex as much as I can, so I came up with this regex-based solution to codingbat.com repeatEnd:

给定一个字符串和一个int N，返回由N个重复的字符串的最后N个字符组成的字符串。您可以假设N介于0和字符串的长度之间（包括0和字符串）。

Given a string and an int N, return a string made of N repetitions of the last N characters of the string. You may assume that N is between 0 and the length of the string, inclusive.

public String repeatEnd(String str, int N) {
  return str.replaceAll(
    ".(?!.{N})(?=.*(?<=(.{N})))|."
      .replace("N", Integer.toString(N)),
    "$1"
  );
}

部分说明：

• 。（？！。{N}）：确认匹配的字符是最后N个字符之一，确保它后面没有N个字符。



• （？=。*（？< =（。{N}）））：在这种情况下，使用lookforward首先一直到字符串的结尾，然后使用嵌套的lookbehind将最后N个字符捕获到 \1 。请注意，此断言将始终为真。

• .(?!.{N}): asserts that the matched character is one of the last N characters, by making sure that there aren't N characters following it.

• (?=.*(?<=(.{N}))): in which case, use lookforward to first go all the way to the end of the string, then a nested lookbehind to capture the last N characters into \1. Note that this assertion will always be true.

|。：如果第一个断言失败（即前面至少有N个字符然后然后匹配字符; \1 将为空。

|.: if the first assertion failed (i.e. there are at least N characters ahead) then match the character anyway; \1 would be empty.

在任何一种情况下，字符始终匹配;用 \1 替换它。

In either case, a character is always matched; replace it with \1.

我的问题是：

• 这种嵌套断言技术是否有效？ （即在前瞻期间向后看？）


• 是否有更简单的基于正则表达式的解决方案？

执行 repeatBegin （如类似定义） 。

Do repeatBegin (as analogously defined).

我老实说这个问题很麻烦！

I'm honestly having troubles with this one!

推荐答案

好一个！虽然我会重构它以避免不必要地使用负逻辑，但我没有看到显着改进该正则表达式的方法：

Nice one! I don't see a way to significantly improve on that regex, although I would refactor it to avoid the needless use of negative logic:

".(?=.{N})|.(?=.*(?<=(.{N})))"

这样，在你到达最终的 N 字符之前，永远不会输入第二种选择，我认为这意味着意图更加清晰。

This way the second alternative is never entered until you reach the final N characters, which I think makes the intent a little clearer.

我从来没有看到一个参考文献说它可以嵌套外观，但就像巴特一样，我不明白为什么它不会。我有时会在lookbehinds中使用lookaheads来绕过可变长度lookbehind表达式的限制。

I've never seen a reference that says it's okay to nest lookarounds, but like Bart, I don't see why it wouldn't be. I sometimes use lookaheads inside lookbehinds to get around limitations on variable-length lookbehind expressions.

编辑：我刚刚意识到我可以通过将更改放在前瞻中来简化正则表达式：

I just realized I can simplify the regex quite a bit by putting the alternation inside the lookahead:

".(?=.{N}|.*(?<=(.{N})))"

顺便说一句，您是否考虑使用 format（）来构建正则表达式而不是 replace（）？

By the way, have you considered using format() to build the regex instead of replace()?

return str.replaceAll(
  String.format(".(?=.{%1$d}|.*(?<=(.{%1$d})))", N),
  "$1"
);

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