问题描述
在广告平台中，为了给广告主一定的自由性和效率，允许广告主在创造标题的时候以通配符的方式进行创意提交。线上服务的时候，会根据用户的搜索词触发的 bidword 对创意中的通配符（通配符是用成对 {} 括起来的字符串，可以包含 0 个或者多个字符）进行替换，用来提升广告投放体验。例如：“{末日血战} 上线送 SSR 英雄，三天集齐无敌阵容！”，会被替换成“帝国时代游戏下载上线送 SSR 英雄，三天集齐无敌阵容！”。给定一个含有通配符的创意和n个标题，判断这句标题是否从该创意替换生成的。
测试样例
样例1：
输入：
n = 4, template = "ad{xyz}cdc{y}f{x}e", titles = ["adcdcefdfeffe", "adcdcefdfeff", "dcdcefdfeffe", "adcdcfe"]
输出："True,False,False,True"
样例2：
输入：
n = 3, template = "a{bdc}efg", titles = ["abcdefg", "abefg", "efg"]
输出："True,True,False"
样例3：
输入：
n = 5, template = "{abc}xyz{def}", titles = ["xyzdef", "abcdef", "abxyzdef", "xyz", "abxyz"]
输出："True,False,True,True,True"
思路
这里关键是需要将中括号里面的内容替换成.*，这样就符合正则表达式的形式，而这一步的替换恰好是正则表达式的匹配问题，即，将{.*?}替换成.*
我们可以使用python的re库
Return the string obtained by replacing the leftmost non-overlapping occurrences of pattern in string by the replacement repl. If the pattern isn’t found, string is returned unchanged. repl can be a string or a function; if it is a string, any backslash escapes in it are processed. That is, \n is converted to a single newline character, \r is converted to a carriage return, and so forth. Unknown escapes of ASCII letters are reserved for future use and treated as errors. Other unknown escapes such as & are left alone. Backreferences, such as \6, are replaced with the substring matched by group 6 in the pattern. For example:
>>> re.sub(r'def\s+([a-zA-Z_][a-zA-Z_0-9]*)\s*(\s*):',
... r'static PyObject*\npy_\1(void)\n{',
... 'def myfunc():')
'static PyObject*\npy_myfunc(void)\n{'
If repl is a function, it is called for every non-overlapping occurrence of pattern. The function takes a single Match argument, and returns the replacement string. For example:
>>> def dashrepl(matchobj):
... if matchobj.group(0) == '-': return ' '
... else: return '-'
...
>>> re.sub('-{1,2}', dashrepl, 'pro----gram-files')
'pro--gram files'
>>> re.sub(r'\sAND\s', ' & ', 'Baked Beans And Spam', flags=re.IGNORECASE)
'Baked Beans & Spam'
Compile a regular expression pattern into a regular expression object, which can be used for matching using its match(), search() and other methods, described below.
The expression’s behaviour can be modified by specifying a flags value. Values can be any of the flags variables, combined using bitwise OR (the | operator).
The sequence
prog = re.compile(pattern)
result = prog.match(string)
is equivalent to
result = re.match(pattern, string)
but using re.compile() and saving the resulting regular expression object for reuse is more efficient when the expression will be used several times in a single program.（由于我们这里要匹配一个list，所以我们先保存正则表达式，然后遍历list进行匹配）
If zero or more characters at the beginning of string match the regular expression pattern, return a corresponding Match. Return None if the string does not match the pattern; note that this is different from a zero-length match.
完整代码
import re
def solution(n, template_, titles):
regex_pattern = re.sub(r'\{.*?\}', '.*', template_)
pattern = re.compile(f'^{regex_pattern}$')
results = []
for title in titles:
if pattern.match(title):
results.append("True")
else:
results.append("False")
return ",".join(results)
if __name__ == "__main__":
# You can add more test cases here
testTitles1 = ["adcdcefdfeffe", "adcdcefdfeff", "dcdcefdfeffe", "adcdcfe"]
testTitles2 = ["CLSomGhcQNvFuzENTAMLCqxBdj", "CLSomNvFuXTASzENTAMLCqxBdj", "CLSomFuXTASzExBdj", "CLSoQNvFuMLCqxBdj", "SovFuXTASzENTAMLCq", "mGhcQNvFuXTASzENTAMLCqx"]
testTitles3 = ["abcdefg", "abefg", "efg"]
print(solution(4, "ad{xyz}cdc{y}f{x}e", testTitles1) == "True,False,False,True" )
print(solution(6, "{xxx}h{cQ}N{vF}u{XTA}S{NTA}MLCq{yyy}", testTitles2) == "False,False,False,False,False,True" )
print(solution(3, "a{bdc}efg", testTitles3) == "True,True,False" )
