【leetcode】1023. Camelcase Matching

题目如下：

A query word matches a given pattern if we can insert lowercaseletters to the pattern word so that it equals the query. (We may insert each character at any position, and may insert 0 characters.)

Given a list of queries, and a pattern, return an answer list of booleans, where answer[i] is true if and only if queries[i] matches the pattern.

 

Example 1:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
Output: [true,false,true,true,false]
Explanation: 
"FooBar" can be generated like this "F" + "oo" + "B" + "ar".
"FootBall" can be generated like this "F" + "oot" + "B" + "all".
"FrameBuffer" can be generated like this "F" + "rame" + "B" + "uffer".

Example 2:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
Output: [true,false,true,false,false]
Explanation: 
"FooBar" can be generated like this "Fo" + "o" + "Ba" + "r".
"FootBall" can be generated like this "Fo" + "ot" + "Ba" + "ll".

Example 3:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
Output: [false,true,false,false,false]
Explanation: 
"FooBarTest" can be generated like this "Fo" + "o" + "Ba" + "r" + "T" + "est".

 

Note:

1. 1 <= queries.length <= 100
2. 1 <= queries[i].length <= 100
3. 1 <= pattern.length <= 100
4. All strings consists only of lower and upper case English letters.

解题思路：这种模式匹配的题目，感觉还是用正则表达式简单些。例如 pattern = "FoBaT"，转换成正则表达式的pattern="^[a-z]*F[a-z]*o[a-z]*B[a-z]*a[a-z]*T[a-z]*$"。

代码如下：

class Solution(object):
    def camelMatch(self, queries, pattern):
        """
        :type queries: List[str]
        :type pattern: str
        :rtype: List[bool]
        """
        res = []
        import re
        rePattern = '^[a-z]*'
        for i,v in enumerate(pattern):
            rePattern += v
            rePattern += '[a-z]*'
        #rePattern = rePattern[:-2]
        rePattern += '$'
        for i in queries:
            r = re.search(rePattern, i)
            if r == None:
                res.append(False)
                continue
            r = r.group()
            subq = i[len(r):]
            res.append(len(subq) == 0 or subq.islower())
        return res