leetcode1004

class Solution:
    def getMax(self,B:'List[int]'):
        n = len(B)
        maxlen = 0
        curlen = 0
        for i in range (n):
            if B[i] == 1:
                curlen += 1
            else:
                maxlen = max(maxlen,curlen)
                curlen = 0
        return max(maxlen,curlen)
    def longestOnes(self, A: 'List[int]', K: int) -> int:
        if K == 0:
            return self.getMax(A)
        n = len(A)
        zlist = list()
        for i in range(n):
            if A[i] == 0:
                zlist.append(i)
        if len(zlist)<=K:
            return n
        maxlen = 0
        for zi in range(len(zlist)-K+1):        
            ti = zi+K
            
            left = 0
            right = len(A)-1
            if zi==0:
                left = zlist[zi]
            else:
                left = zlist[zi-1]+1
            if zi == len(zlist)-K:
                right = len(A) - 1
            else:
                right = zlist[ti] - 1

            maxlen = max(maxlen, right - left +1)
            
        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。