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?
Solution：滑动窗[l, h]
思路：
这道题在之前那道题Max Consecutive Ones的基础上加了一个条件，有一次将0翻转成1的机会，问此时最大连续1的个数。
用一个变量cnt来记录连续1的个数，当遇到了0的时候，因为我们有一次0变1的机会，所以我们遇到0了还是要累加cnt，然后我们此时需要用另外一个变量cur来保存当前cnt的值，然后cnt重置为0，以便于让cnt一直用来统计纯连续1的个数，然后我们每次都用用cnt+cur来更新结果res。
Time Complexity: O(N) Space Complexity: O(1)
Solution Code：
class Solution {
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;
}
}
