假设我有(在
javascript正则表达式中)
((((A)B)C)D)
当然,这真的是读
ABCD
是否有算法消除字符串中不必要的括号?
最佳答案 此函数将删除未跟随量词的所有组,并且不是环顾四周.它假设ECMAScript风格正则表达式,捕获组((…))是不重要的.
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("");
}
例子:
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只包含一个标记((A)?).这将需要更长的标记化模式.