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1023.py
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1023.py
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'''
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.
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.
'''
class Solution(object):
def camelMatch(self, queries, pattern):
"""
:type queries: List[str]
:type pattern: str
:rtype: List[bool]
"""
import re
result = []
patterns = re.findall('[A-Z][a-z]*', pattern)
for query in queries:
splitter = re.findall('[A-Z][a-z]*', query)
flag = True
if len(patterns) == len(splitter):
for index in range(len(patterns)):
# print patterns[index], splitter[index]
p_i, s_i = 1, 1
if patterns[index][0] == splitter[index][0]:
while p_i < len(patterns[index]) and s_i < len(splitter[index]):
if patterns[index][p_i] == splitter[index][s_i]:
p_i += 1
s_i += 1
else:
s_i += 1
if p_i != len(patterns[index]):
flag = False
break
else:
flag = False
break
if flag:
result.append(True)
else:
result.append(False)
else:
result.append(False)
return result