我想从数值范围生成(系列)regexp(s).
例：
1013 - 4044 => regexp matches --------------------------------------- 101[3-9] 1013 - 1019 10[2-9][0-9] 1020 - 1099 11[0-9][0-9] 1100 - 1199 [23][0-9][0-9][0-9] 2000 - 3999 40[0-3][0-9] 4000 - 4039 404[0-4] 4040 - 4044
什么是最简单的算法？
扭转它的最简单方法是什么(即给出正则表达式,寻找范围)？
很高兴看到java,clojure,perl中的解决方案……
谢谢！
给定范围有一个
online tool用于生成正则表达式,并提供解释.您也可以在那里找到源代码.例如：
^(101[3-9]|10[2-9][0-9]|1[1-9][0-9]{2}|[23][0-9]{3}|40[0-3][0-9]|404[0-4])$
First,break into equal length ranges: 1013 - 4044 Second,break into ranges that yield simple regexes: 1013 - 1019 1020 - 1099 1100 - 1999 2000 - 3999 4000 - 4039 4040 - 4044 Turn each range into a regex: 101[3-9] 10[2-9][0-9] 1[1-9][0-9]{2} [23][0-9]{3} 40[0-3][0-9] 404[0-4] Collapse adjacent powers of 10: 101[3-9] 10[2-9][0-9] 1[1-9][0-9]{2} [23][0-9]{3} 40[0-3][0-9] 404[0-4] Combining the regexes above yields: (101[3-9]|10[2-9][0-9]|1[1-9][0-9]{2}|[23][0-9]{3}|40[0-3][0-9]|404[0-4]) Next we'll try factoring out common prefixes using a tree: Parse into tree based on regex prefixes: . 1 0 1 [3-9] + [2-9] [0-9] + [1-9] [0-9]{2} + [23] [0-9]{3} + 4 0 [0-3] [0-9] + 4 [0-4] Turning the parse tree into a regex yields: (1(0(1[3-9]|[2-9][0-9])|[1-9][0-9]{2})|[23][0-9]{3}|40([0-3][0-9]|4[0-4])) We choose the shorter one as our result. ^(101[3-9]|10[2-9][0-9]|1[1-9][0-9]{2}|[23][0-9]{3}|40[0-3][0-9]|404[0-4])$
要反转它,您可以查看字符类,并获得每个替代项的最小值和最大值.
^(101[3-9]|10[2-9][0-9]|1[1-9][0-9]{2}|[23][0-9]{3}|40[0-3][0-9]|404[0-4])$
=> 1013 1020 1100 2000 4000 4040 lowers
1019 1999 1199 3999 4039 4044 uppers
=> 1013 - 4044
