LeetCode One question per day 1004. The maximum number of consecutive 1s III

1004. Maximum number of consecutive 1s III

Given an array A consisting of several 0s and 1, we can change up to K values ​​from 0 to 1.

Returns the length of the longest (contiguous) sub-array containing only 1.

Example 1:

输入：A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
输出：6
解释： 
[1,1,1,0,0,1,1,1,1,1,1]
粗体数字从 0 翻转到 1，最长的子数组长度为 6。

Example 2:

输入：A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3
输出：10
解释：
[0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
粗体数字从 0 翻转到 1，最长的子数组长度为 10。

prompt:

• 1 <= A.length <= 20000
• 0 <= K <= A.length
• A[i] is 0 or 1

Method one: sliding window

Problem-solving ideas

• Double pointer still [left, right)sliding window simulation
• rightTraverse the pointer to the right, when encountering Ka 0time, the value of the calculation result ret = max(ret, right - left)and moves leftthe pointer to the next 0right .

Reference Code

public int longestOnes(int[] A, int K) {
    
    
    int n = A.length;
    int ret = 0;
    int left = 0, right = 0, count = 0;
    while (right < n) {
    
    
        if (A[right] == 0) {
    
    
            if (count < K) {
    
    
                count++;
            } else {
    
    
                ret = Math.max(ret, right - left);
                do {
    
    
                    left++;
                } while (A[left - 1] == 1);
            }
        }
        right++;
    }
    return Math.max(ret, right - left);
}

Results of the

• Time complexity: O(n)
• Space complexity: O(1)