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

问题描述

这与我之前的工作类似( wordEnds 和 repeatEnd ):作为一项心理锻炼，我想解决这个问题仅使用正则表达式的玩具问题.

This is similar to my previous efforts (wordEnds and repeatEnd): as a mental exercise, I want to solve this toy problem using regex only.

codingbat.com的描述:

给出一个字符串和一个非空的单词字符串，返回原始字符串的版本，其中所有字符都被加号("+" )代替，除了单词字符串的外观保留不变.

Given a string and a non-empty word string, return a version of the original string where all chars have been replaced by pluses ("+"), except for appearances of the word string which are preserved unchanged.

plusOut("12xy34", "xy") → "++xy++"
plusOut("12xy34", "1") → "1+++++"
plusOut("12xy34xyabcxy", "xy") → "++xy++xy+++xy"

没有提及是否允许重叠(例如，什么是 plusOut("+ xAxAx +"，"xAx")?)，但是我的非正则表达式解决方案无法处理重叠和它通过了，所以我想如果它使 word 变得更简单，我们可以假设它不重叠(如果您提供两种变体的解决方案，都可以加分！).

There is no mention whether or not to allow overlap (e.g. what is plusOut("+xAxAx+", "xAx")?), but my non-regex solution doesn't handle overlap and it passes, so I guess we can assume non-overlapping occurrences of word if it makes it simpler (bonus points if you provide solutions for both variants!).

无论如何，我都想使用正则表达式来解决这个问题(与以前在其他两个问题上一样).我什至没有任何可显示的东西，因为我什么都没用.

In any case, I'd like to solve this using regex (of the same style that I did before with the other two problems), but I'm absolutely stumped. I don't even have anything to show, because I have nothing that works.

因此，让我们看看stackoverflow社区带来了什么.

So let's see what the stackoverflow community comes up with.

推荐答案

这通过了所有测试:

public String plusOut(String str, String word) {
  return str.replaceAll(
    String.format("(?<!(?=\\Q%s\\E).{0,%d}).", word, word.length()-1),
    "+"
  );  
}

我也得到:

plusOut("1xAxAx2", "xAx") → "+xAxAx+"

如果这是您想要的结果，那么我也通过了重叠测试，但是我不得不承认，这是偶然的.:D

If that's the result you were looking for then I pass your overlap test as well, but I have to admit, that one's by accident. :D

这篇关于使用正则表达式的codingBat plusOut的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！