Description
A query word matches a given pattern if we can insert lowercase letters 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.
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。
题目链接：https://leetcode.com/problems/camelcase-matching/
Difficulty: medium
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 <= queries.length <= 100
- 1 <= queries[i].length <= 100
- 1 <= pattern.length <= 100
- All strings consists only of lower and upper case English letters.
分析
- 对每一个query做如下操作：
- 遍历pattern，比较query和pattern，若不等，且当前query为大写字母，退出，匹配失败；
- 反正继续向后匹配，若匹配玩query，但没有匹配玩pattern，失败；
- 匹配完pattern，若生于query没有大写字母，匹配成功，反之，失败；
- 结果存储在res，返回res。
参考代码
class Solution(object):
def camelMatch(self, queries, pattern):
ans=[]
for query in queries:
i=0
judge=True
for p in pattern:
while(judge and i < len(query) and p != query[i]):
if(query[i].isupper()):
judge=False
ans.append(False)
i=len(query)+1
break
i+=1
if(i==len(query)):
ans.append(False)
break
i+=1
else:
while(i<len(query)):
if(query[i].isupper()):
ans.append(False)
break
else:
i+=1
if(i==len(query)):
ans.append(True)
return ans