# LeetCode: 1023. 驼峰式匹配¶

## 1、题目描述¶

输入：queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"

"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".


输入：queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"

"FooBar" 可以这样生成："Fo" + "o" + "Ba" + "r".
"FootBall" 可以这样生成："Fo" + "ot" + "Ba" + "ll".


输出：queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"

"FooBarTest" 可以这样生成："Fo" + "o" + "Ba" + "r" + "T" + "est".


• 1 <= queries.length <= 100
• 1 <= queries[i].length <= 100
• 1 <= pattern.length <= 100
• 所有字符串都仅由大写和小写英文字母组成。

## 2、解题思路¶

• 直接匹配出所有的Partten，判断不能匹配的字符中是否有大写字母
from string import ascii_uppercase

class Solution:
def camelMatch(self, queries: List[str], pattern: str) -> List[bool]:
upper = set(ascii_uppercase)

def judge(word, part):
w_pos = 0
p_pos = 0
not_equal = set()
while p_pos < len(part) and w_pos < len(word):
if word[w_pos] == part[p_pos]:
p_pos += 1
else: