删除正则表达式中不必要的括号

发布于 2024-09-04 04:24:39 字数 151 浏览 6 评论 0原文

假设我（在 JavaScript 正则表达式中）

((((A)B)C)D)

当然是这样的：

ABCD

是否有一种算法可以消除这样的字符串中不必要的括号？

原文

Suppose I have (in a javascript regular expression)

((((A)B)C)D)

Of course that really reads

ABCD

Is there an algorithm to eliminate unnecessary parentheses in a string like that?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

云柯 2024-09-11 04:24:39

此函数将删除所有后面没有量词的组，并且不是环视。它假定 ECMAScript 风格的正则表达式，并且捕获组 (( ... )) 并不重要。

function removeUnnecessaryParenthesis(s) {
   // Tokenize the pattern
   var pieces = s.split(/(\\.|\[(?:\\.|[^\]\\])+]|\((?:\?[:!=])?|\)(?:[*?+]\??|\{\d+,?\d*}\??)?)/g);
   var stack = [];
   for (var i = 0; i < pieces.length; i++) {
      if (pieces[i].substr(0,1) == "(") {
         // Opening parenthesis
         stack.push(i);
      } else if (pieces[i].substr(0,1) == ")") {
         // Closing parenthesis
         if (stack.length == 0) {
            // Unbalanced; Just skip the next one.
            continue;
         }
         var j = stack.pop();
         if ((pieces[j] == "(" || pieces[j] == "(?:") && pieces[i] == ")") {
             // If it is a capturing group, or a non-capturing group, and is
             // not followed by a quantifier;
             // Clear both the opening and closing pieces.
             pieces[i] = "";
             pieces[j] = "";
         }
      }
   }
   return pieces.join("");
}

示例：

removeUnnecessaryParenthesis("((((A)B)C)D)")  --> "ABCD"
removeUnnecessaryParenthesis("((((A)?B)C)D)") --> "(A)?BCD"
removeUnnecessaryParenthesis("((((A)B)?C)D)") --> "(AB)?CD"

它不会尝试确定括号是否仅包含单个标记 ((A)?)。这将需要更长的标记化模式。

This function will remove all groups that is not followed by a quantifier, and is not a look-around. It assumes ECMAScript flavor regex, and that capture-groups (( ... )) are unimportant.

function removeUnnecessaryParenthesis(s) {
   // Tokenize the pattern
   var pieces = s.split(/(\\.|\[(?:\\.|[^\]\\])+]|\((?:\?[:!=])?|\)(?:[*?+]\??|\{\d+,?\d*}\??)?)/g);
   var stack = [];
   for (var i = 0; i < pieces.length; i++) {
      if (pieces[i].substr(0,1) == "(") {
         // Opening parenthesis
         stack.push(i);
      } else if (pieces[i].substr(0,1) == ")") {
         // Closing parenthesis
         if (stack.length == 0) {
            // Unbalanced; Just skip the next one.
            continue;
         }
         var j = stack.pop();
         if ((pieces[j] == "(" || pieces[j] == "(?:") && pieces[i] == ")") {
             // If it is a capturing group, or a non-capturing group, and is
             // not followed by a quantifier;
             // Clear both the opening and closing pieces.
             pieces[i] = "";
             pieces[j] = "";
         }
      }
   }
   return pieces.join("");
}

Examples:

removeUnnecessaryParenthesis("((((A)B)C)D)")  --> "ABCD"
removeUnnecessaryParenthesis("((((A)?B)C)D)") --> "(A)?BCD"
removeUnnecessaryParenthesis("((((A)B)?C)D)") --> "(AB)?CD"

It does not try to determine if the parenthesis contains only a single token ((A)?). That would require a longer tokenizing pattern.

回复收藏 0 原文