LeetCode 1004. Max Consecutive Ones III

原题链接在这里：https://leetcode.com/problems/max-consecutive-ones-iii/

题目：

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 

题解：

Same idea as Max Consecutive Ones II.

Solution is easy to extend from 1 to K.

Time Complexity: O(n). n = A.length.

Space: O(1).

AC Java:

class Solution {
    public int longestOnes(int[] A, int K) {
        if(A == null || A.length == 0){
            return 0;
        }
        
        int count = 0;
        int walker = 0;
        int runner = 0;
        int res = 0;
        while(runner < A.length){
            if(A[runner++] != 1){
                count++;
            }
            
            while(count > K){
                if(A[walker++] != 1){
                    count--;
                }
            }
            
            res = Math.max(res, runner-walker);
        }
        
        return res;
    }
}