題目如下：

Given an array A of 0s and 1s, we may change up to K values from 0 to 1.

Return the length of the longest (contiguous) subarray that contains only 1s.

Example 1:

Input: A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
Output: 6
Explanation: 
[1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1.  The longest subarray is underlined.
 

Example 2:

Input: A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3
Output: 10
Explanation: 
[0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1.  The longest subarray is underlined.

Note:

1. 1 <= A.length <= 20000
2. 0 <= K <= A.length
3. A[i] is 0 or 1

解題思路：如果我們把所有0所在位置對應的下標存入一個數組inx_0中，例如Example 2的inx_0 = [0,1,4,5,9,12,13,14]，因為最多把K個0變成1，要構造出最長的全為1的連續子數組，那麽很顯然所有進行反轉操作的0所對應下標在inx_0中必須是連續的。接下來開始遍歷數組A，在K大於0的條件下，遇到1則count_1加1，遇到0的話則K減1並且count_1加1（表示這個0反轉成1）；如果在K=0的情況下遇到0，那麽需要去掉第一個0變成的1和這個0之前連續的1的數量，即count減1（減去第一個0變成1的計數），再減去第一個0前面連續的1的數量i，記為count_1減i。記錄count_1出現的最大值，直到數組A遍歷完成為止。

代碼如下：

class Solution(object):
    def longestOnes(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        count_1 = 0
        zero_inx = []
        res = 0
        zero_before = -1
        for i in range(len(A)):
            if A[i] == 1:
                count_1 
 
+= 1
            else:
                zero_inx.append(i)
                if K > 0:
                    K -= 1
                    count_1 += 1
                elif K == 0:
                    res = max(res,count_1)
                    first_0 = zero_inx.pop(0)
                    count_1 -= (first_0 - zero_before - 1)
                    zero_before = first_0
        res = max(res, count_1)
        return res

【leetcode】1004. Max Consecutive Ones III