考卷的所有題目的答案用一列T、F組成的字符串表示，可以操作k次，將T改成F或?qū)改成T，問(wèn)最多可以制造多少個(gè)連續(xù)的T或者F
思路參考 -> https://leetcode.com/problems/maximize-the-confusion-of-an-exam/solutions/3729656/video-solution-sliding-window-2-pointers/
維護(hù)兩個(gè)游標(biāo)i和j，j走在前，當(dāng)i～j之間T和F之中最少出現(xiàn)的那一個(gè)超過(guò)k次時(shí)，將這一區(qū)間的統(tǒng)一改為T(mén)或者F
1 #2024
2 #Runtime: 298 ms (Beats 84.38%)
3 #Memory: 13.6 MB (Beats 96.88%)
4
5 class Solution(object):
6 def maxConsecutiveAnswers(self, answerKey, k):
7 """
8 :type answerKey: str
9 :type k: int
10 :rtype: int
11 """
12 cnt_F, cnt_T = 0, 0
13 i, j = 0, 0
14 ans = 0
15 while j < len(answerKey):
16 if answerKey[j] == 'F':
17 cnt_F += 1
18 else:
19 cnt_T += 1
20 while min(cnt_T, cnt_F) > k:
21 if answerKey[i] == 'F':
22 cnt_F -= 1
23 else:
24 cnt_T -= 1
25 i += 1
26 ans = max(ans, cnt_F + cnt_T)
27 j += 1
28 return ans