有效地找到正则表达式的所有重叠匹配

时间:2012-07-07 00:43:10

标签: java regex optimization

这是All overlapping substrings matching a java regex的后续内容。

有没有办法让这段代码更快?

public static void allMatches(String text, String regex)
  {
    for (int i = 0; i < text.length(); ++i) {
      for (int j = i + 1; j <= text.length(); ++j) {
        String positionSpecificPattern = "((?<=^.{"+i+"})("+regex+")(?=.{"+(text.length() - j)+"}$))";
        Matcher m = Pattern.compile(positionSpecificPattern).matcher(text);

        if (m.find()) 
        {   
          System.out.println("Match found: \"" + (m.group()) + "\" at position [" + i + ", " + j + ")");
        }   
      }   
    }   
  }

1 个答案:

答案 0 :(得分:1)

在另一个问题中,您提到了Matcher的region()方法,但您没有充分利用它。使其如此有价值的是锚点将在区域的边界处匹配,就好像它们是独立字符串的边界一样。假设您已设置useAnchoringBounds()选项,但这是默认设置。

public static void allMatches(String text, String regex)
{
  Matcher m = Pattern.compile(regex).matcher(text);
  int end = text.length();
  for (int i = 0; i < end; ++i)
  {
    for (int j = i + 1; j <= end; ++j) 
    {
      m.region(i, j);

      if (m.find()) 
      {   
        System.out.printf("Match found: \"%s\" at position [%d, %d)%n",
                          m.group(), i, j);
      }   
    }   
  }   
}

给出你的样本字符串和正则表达式:

allMatches("String t = 04/31 412-555-1235;", "^\\d\\d+$");

...我得到了这个输出:

Match found: "04" at position [11, 13)
Match found: "31" at position [14, 16)
Match found: "41" at position [17, 19)
Match found: "412" at position [17, 20)
Match found: "12" at position [18, 20)
Match found: "55" at position [21, 23)
Match found: "555" at position [21, 24)
Match found: "55" at position [22, 24)
Match found: "12" at position [25, 27)
Match found: "123" at position [25, 28)
Match found: "1235" at position [25, 29)
Match found: "23" at position [26, 28)
Match found: "235" at position [26, 29)
Match found: "35" at position [27, 29)