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;
     }
 }