487. Max Consecutive Ones II

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本文探讨了在一个二进制数组中找到最多连续1的问题，允许翻转一个0。采用双指针技巧，高效地求解该问题，并提供了一个简洁的Java实现方案。

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?

用双指针解题，一个快的指针，从前向后依次遍历，遇到0，就记录之前1的长度，同时设置慢指针到当前位置，每次比较（快指针-满指针+之前1的长度）和max的大小，取较大的。代码如下：

public class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int slower = -1, pre1s = 0, max = 0;
        for (int i = 0; i < nums.length; i ++) {
            if (nums[i] == 0) {
                pre1s = i - slower;
                slower = i;
            }
            max = Math.max(max, i - slower + pre1s);
        }
        return max;
    }
}