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.

分析：
用两个指针: start, end. 不断往前移动end，当arr[end]是0的话，就增加count的个数。如果count的个数超过k，那么就往前移动start, 然后不断比较global_max和 end - start + 1.

 1 class Solution {
 2     public int longestOnes(int[] A, int K) {
 3         int zeroCount = 0, start = 0, res = 0;
 4         for (int end = 0; end < A.length; end++) {
 5             if (A[end] == 0) {
 6                 zeroCount++;
 7             }
 8             while (zeroCount > K) {
 9                 if (A[start] == 0) {
10                     zeroCount--;
11                 }
12                 start++;
13             }
14             res = Math.max(res, end - start + 1);
15         }
16         return res;
17     }
18 }