1004. 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 

class Solution {
    public int longestOnes(int[] arr, int k) {
        int res = 0, start = 0;
        int ocount = 0;
        
        for(int end = 0; end < arr.length; end++) {
            if(arr[end] == 1) ocount++;
            while(end - start + 1 - ocount > k) {
                if(arr[start] == 1) ocount--;
                start++;                
            }
            res = Math.max(res, end - start + 1);
        }
        return res;
    }
}

这题也是sliding window，是上面那道题的简化版，window维护啥呢？维护在k次改变下能生成的最长1子串。上个题有26个字母，这个只有0和1，而且要求要是1子串

开始，如果当前数字==1，onecount++，如果当前window包不住了，就要start++，注意啊如果arr【start】==1了菜肴onecount--。