java - 正则表达式不给出溢出错误

Question

示例代码:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Regex {
    public static void main(String[] args) {
        String data = "Shyam and you. You are 2.3 km away from home. Lakshmi and you. Ram and you. You are Mike. ";
        Pattern pattern = Pattern.compile(
                "\\s*((?:[^\\.]|(?:\\w+\\.)+\\w)*are.*?)(?:\\.\\s|\\.$)",
                Pattern.DOTALL);
        Matcher matcher = pattern.matcher(data);
        while (matcher.find()) {
            System.out.println(matcher.group(0));
        }
    }
}

输出:

You are 2.3 km away from home. 

You are Mike. 

我在执行上面的代码时得到了预期的输出。 但问题是当用更大的字符串测试同一个正则表达式时，它显示溢出错误。 我进行了大致相同的搜索，发现正则表达式中的 (A|B)* 之类的交替会导致问题。 有什么办法可以解决这个问题吗？ 请帮忙。

Answer 1

我已尝试重构您的正则表达式以避免回溯。你能试试这个正则表达式吗:

Pattern pattern = Pattern.compile("(?>[^.]|(?:\\w+\\.)+\\w)+\\sare\\s.*?(?>\\.\\s|\\.$)",
                  Pattern.DOTALL);

(?>group) 称为原子分组。

根据:http://www.regular-expressions.info/atomic.html

原子分组

An atomic group is a group that, when the regex engine exits from it, automatically throws away all backtracking positions remembered by any tokens inside the group. Atomic groups are non-capturing. The syntax is (?>group).