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 

思路：典型的同向双指针题目，count最长的区间使得最多含有K个0；O(N);

class Solution {
    public int longestOnes(int[] A, int K) {
        if(A == null || A.length == 0 || K < 0) {
            return 0;
        }
        int maxlen = 0;
        int count = 0;
        int j = 0;
        // two pointers algorithm;
        for(int i = 0; i < A.length; i++) {
            // move j;
            while(j < A.length && count <= K) {
                if(A[j] == 0) {
                    if(count == K) {
                        break;
                    }
                    count++;
                }
                j++;
            }
            
            // update result;
            maxlen = Math.max(maxlen, j - i);
            
            // move i;
            if(A[i] == 0) {
                count--;
            }
        }
        return maxlen;
    }
}