Max Consecutive Ones III

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 

题目理解：

给定一个由0和1组成的数组，将其中的某K个0变成1之后，数组中存在的最长的连续的1有多长

解题思路：

用dp[i]表示到i位置为止，有多少个0，使用两个指针left和right，每次固定left，移动right，保证right - left <= K，记录最大的（right - left）。然后将left加1，继续移动right。right可以一直增加，因为如果right变小，那么（right - left）一定小于已经记录的最大值。

class Solution {
    public int longestOnes(int[] A, int K) {
        int len = A.length;
        int[] record = new int[len + 1];
        for(int i = 1; i < len + 1; i++){
            record[i] = record[i - 1];
            if(A[i - 1] == 0)
                record[i]++;
        }
        int left = 0, right = 0, res = 0;
        while(true){
            while(right < len && record[right + 1] - record[left] <= K)
                right++;
            res = Math.max(res, right - left);
            if(right == len)
                break;
            left++;
        }
        return res;
    }
}