1004. Max Consecutive Ones III

置顶

Created At : 2023-10-28 15:14

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

这道题是之前那道 Max Consecutive Ones II 的拓展，博主在之前的解法二中就预言了翻转k次的情况，果不其然，在五百多道题之后，该来的还是来了。不过不要紧，我们已经知道了该如何应对了，直接上滑动窗口 Sliding Window，用个变量 cnt 记录当前将0变为1的个数，在遍历数组的时候，若遇到了0，则 cnt 自增1。若此时 cnt 大于K了，说明该缩小窗口了，用个 while 循环，若左边界为0，移除之后，此时 cnt 应该自减1，left 自增1，每次用窗口大小更新结果 res 即可，参见代码如下：

解法一：

class Solution {
public:
    int longestOnes(vector<int>& A, int K) {
        int n = A.size(), left = 0, cnt = 0, res = 0;
        for (int i = 0; i < n; ++i) {
            if (A[i] == 0) ++cnt;
            while (cnt > K) {
                if (A[left] == 0) --cnt;
                ++left;
            }
            res = max(res, i - left + 1);
        }
        return res;
    }
};

我们也可以写的更简洁一些，不用 while 循环，但是还是用的滑动窗口的思路，其中i表示左边界，j为右边界。在遍历的时候，若遇到0，则K自减1，若K小于0了，且 A[i] 为0，则K自增1，且i自增1，最后返回窗口的大小即可，参见代码如下：

解法二：

class Solution {
public:
    int longestOnes(vector<int>& A, int K) {
        int n = A.size(), i = 0, j = 0;
        for (; j < n; ++j) {
            if (A[j] == 0) --K;
            if (K < 0 && A[i++] == 0) ++K;
        }
        return j - i;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1004

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