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