问题描述
我想找到所有可能的正则表达式匹配，这怎么可能?
I would like to find all possible matches of regex, how is it possible?
regex rx("(2|25)");
string s = "2225";
for (sregex_iterator it(s.begin(), s.end(), rx), end; it != end; ++it) {
cout << it->position() << ": " << it->str() << endl;
}
给出输出:
0: 2
1: 2
2: 25
但是找不到确切的第三个 2:2 .我更喜欢使用正则表达式，因为 O(n)的复杂性用于同时搜索多个令牌.
But can't find third 2: 2 exactly. I prefer to use regex because of O(n) complexity for searching several tokens at same time.
更新:
也许将令牌列表拆分为非前缀列表并创建多个正则表达式?例如:(2 | 4 | 25 | 45 | 251 | 455 | 267) => (2 | 4)，(25 | 45 | 267)，(251 | 455)，这将使复杂性增加到 O(n log(m))
Maybe split token list to non-prefixable lists and create several regexes? For example: (2|4|25|45|251|455|267) => (2|4), (25|45|267), (251|455) This will grow complexity to something like O(n log(m))
更新2:
请提供基于STL的简短算法，将令牌向量拆分为非前缀向量，以回答此问题.
Please, provide short STL-based algorithm of splitting token vector to non-prefixable vectors to answer this question.
推荐答案
我认为使用迭代器和单个正则表达式是不可能的.这是它的工作方式.
I dont think it's possible with an iterator and a single regexp. Here's how it works.
您的正则表达式搜索子字符串为"2" 或"25".现在，您可以使用 sregex_iterator 开始搜索.它以字符串的第一个符号开头，并尝试查找与您的正则表达式匹配的内容.如果存在匹配项，则将其记录"，并将迭代器前进到匹配项之后的位置.如果没有匹配项，则迭代器将前进1个位置.这个过程一直持续到到达字符串末尾为止.
Your regexp searches for a substring that is either "2" or "25". Now, you start the search with sregex_iterator. It starts with the first symbol of the string, and tries to find match with your regular expression. If there is a match, it is "recorded", and the iterator is advanced to the position after the match. If there is no match, the iterator is advanced 1 position forward. This process continues until the end of the string is reached.
现在，每次找到匹配项时，它都会尝试从您的正则表达式中找到最佳(即最长)匹配项.因此，如果子字符串同时匹配 2 和 25 ，则它将花费 25 ，因为它更长.所以我说你需要2个正则表达式.
Now, each time it finds a match it will try to find the best (i.e., longest) match from your regular expression. So if a substring matches both 2 and 25, it will take 25 since it's longer. So I'd say you need 2 regular expressions.
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