Leetcode: Max Consecutive Ones II(unsolved locked problem)

Given a binary array, find the maximum number of consecutive 1s in this array if you can flip at most one 0.

Example 1:
Input: [1,0,1,1,0]
Output: 4
Explanation: Flip the first zero will get the the maximum number of consecutive 1s.
    After flipping, the maximum number of consecutive 1s is 4.
Note:

The input array will only contain 0 and 1.
The length of input array is a positive integer and will not exceed 10,000
Follow up:
What if the input numbers come in one by one as an infinite stream? In other words, you can't store all numbers coming from the stream as it's too large to hold in memory. Could you solve it efficiently?

未研究：

The idea is to keep a window [l, h] that contains at most k zero

The following solution does not handle follow-up, because nums[l] will need to access previous input stream
Time: O(n) Space: O(1)

    public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0, zero = 0, k = 1; // flip at most k zero
        for (int l = 0, h = 0; h < nums.length; h++) {
            if (nums[h] == 0)
                zero++;
            while (zero > k)
                if (nums[l++] == 0)
                    zero--;
            max = Math.max(max, h - l + 1);
        }
        return max;
    }

Now let's deal with follow-up, we need to store up to k indexes of zero within the window [l, h] so that we know where to move lnext when the window contains more than k zero. If the input stream is infinite, then the output could be extremely large because there could be super long consecutive ones. In that case we can use BigInteger for all indexes. For simplicity, here we will use int
Time: O(n) Space: O(k)

    public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0, k = 1; // flip at most k zero
        Queue<Integer> zeroIndex = new LinkedList<>();
        for (int l = 0, h = 0; h < nums.length; h++) {
            if (nums[h] == 0)
                zeroIndex.offer(h);
            if (zeroIndex.size() > k)
                l = zeroIndex.poll() + 1;
            max = Math.max(max, h - l + 1);
        }
        return max;
    }

Note that setting k = 0 will give a solution to the earlier version Max Consecutive Ones

For k = 1 we can apply the same idea to simplify the solution. Here q stores the index of zero within the window [l, h] so its role is similar to Queue in the above solution

    public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0, q = -1;
        for (int l = 0, h = 0; h < nums.length; h++) {
            if (nums[h] == 0) {
                l = q + 1;
                q = h;
            }
            max = Math.max(max, h - l + 1);
        }
        return max;
    }