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Q:

如果我们可以将

小写字母

插入模式串

pattern

得到待查询项

query

，那么待查询项与给定模式串匹配。（我们可以在任何位置插入每个字符，也可以插入 0 个字符。）

给定待查询列表

queries

，和模式串

pattern

，返回由布尔值组成的答案列表

answer

。只有在待查项

queries[i]

与模式串

pattern

匹配时，

answer[i]

才为

true

，否则为

false

。

示例 1：

输入：queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
输出：[true,false,true,true,false]
示例：
"FooBar" 可以这样生成："F" + "oo" + "B" + "ar"。
"FootBall" 可以这样生成："F" + "oot" + "B" + "all".
"FrameBuffer" 可以这样生成："F" + "rame" + "B" + "uffer".

连接：

https://leetcode-cn.com/problems/camelcase-matching/

思路：写的不好 乱糟糟

代码：

class Solution:
    def camelMatch(self, queries, pattern):
        res = ['#' for _ in range(len(queries))]
        for s in pattern:
            for i in range(len(queries)):
                if pattern[-1] in queries[i]:
                    end = len(queries[i]) - ''.join(list(queries[i])[::-1]).index(pattern[-1])
                    if s in queries[i] and queries[i].index(s) < end:
                        tmp = (queries[i])
                        tmp = (tmp.replace(str(s), '', 1))
                        queries[i] = tmp
                    else:
                        res[i] = False
                else:
                    res[i] = False

        for i in range(len(queries)):
            for w_ in queries[i]:
                if w_ >= 'A' and w_ < 'Z':
                    res[i] = False
                    break
            if res[i] == '#':
                res[i] = (True)
        return res