java 编写正则表达式来检测重复字符
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Writing a regex to detect repeat-characters
提问by sharon Hwk
I need to write a regex, that would identify a word that have a repeating character setat the end. According to the following code fragment, the repeating character set is An. I need to write a regex so this will be spotted and displayed.
我需要编写一个正则表达式，它可以识别一个在末尾具有重复字符集的单词。根据以下代码片段，重复字符集为. 我需要编写一个正则表达式，以便发现并显示它。An
According to the following code, \\wwill match any word character (including digit, letter, or special character). But i only want to identify english characters.
根据以下代码，\\w将匹配任何单词字符（包括数字、字母或特殊字符）。但我只想识别英文字符。
String stringToMatch = "IranAnAn";
Pattern p = Pattern.compile("(\w)\1+");
Matcher m = p.matcher(stringToMatch);
if (m.find())
{
System.out.println("Word contains duplicate characters " + m.group(1));
}
UPDATE
更新
Word contains duplicate characters a
Word contains duplicate characters a
Word contains duplicate characters An
采纳答案by assylias
You want to catch as many characters in your set as possible, so instead of (\\w)you should use (\\w+)and you want the sequence to be at the end, so you need to add $(and I have removed the +after \\1which is not useful to detect repetition: only one repetition is needed):
你想在你的集合中捕捉尽可能多的字符，所以(\\w)你应该使用而不是(\\w+)你想要序列在最后，所以你需要添加$（我已经删除了对检测重复没有用的+后面\\1：只需要重复一次）：
Pattern p = Pattern.compile("(\w+)\1$");
Your program then outputs Anas expected.
然后您的程序An按预期输出。
Finally, if you only want to capture ascii characters, you can use [a-zA-Z]instead of \\w:
最后，如果您只想捕获 ascii 字符，您可以使用[a-zA-Z]代替\\w：
Pattern p = Pattern.compile("([a-zA-Z]+)\1$");
And if you want the character set to be at least 2 characters:
如果您希望字符集至少为 2 个字符：
Pattern p = Pattern.compile("([a-zA-Z]{2,})\1$");
回答by Michael Lang
If by "only English characters" you mean A-Z and a-z, the follow regex will work:
如果“仅英文字符”是指 AZ 和 az，则以下正则表达式将起作用：
".*([A-Za-z]{2,})\1$"
