leetcode1004

 1 class Solution:
 2     def getMax(self,B:'List[int]'):
 3         n = len(B)
 4         maxlen = 0
 5         curlen = 0
 6         for i in range (n):
 7             if B[i] == 1:
 8                 curlen += 1
 9             else:
10                 maxlen = max(maxlen,curlen)
11                 curlen = 0
12         return max(maxlen,curlen)
13     def longestOnes(self, A: 'List[int]', K: int) -> int:
14         if K == 0:
15             return self.getMax(A)
16         n = len(A)
17         zlist = list()
18         for i in range(n):
19             if A[i] == 0:
20                 zlist.append(i)
21         if len(zlist)<=K:
22             return n
23         maxlen = 0
24         for zi in range(len(zlist)-K+1):        
25             ti = zi+K
26             
27             left = 0
28             right = len(A)-1
29             if zi==0:
30                 left = zlist[zi]
31             else:
32                 left = zlist[zi-1]+1
33             if zi == len(zlist)-K:
34                 right = len(A) - 1
35             else:
36                 right = zlist[ti] - 1
37 
38             maxlen = max(maxlen, right - left +1)
39             
40         return maxlen

经过了几次尝试，终于作出来了。主要的思路是滑动窗口：

先记录所有的0的索引，然后选择等K宽的窗口，计算窗口“所连接”的连续1的起止坐标。然后滑动窗口，进行比较，保留最大值。

posted on 2019-03-04 14:08 Sempron2800+ 阅读(139) 评论(0) 编辑收藏举报