Given an n x n matrix where each of the rows and columns is sorted in ascending order, return the kth smallest element in the matrix.
Note that it is the kth smallest element in the sorted order, not the kth distinct element.
You must find a solution with a memory complexity better than O(n2).
\n
Example 1:
\n\n\nInput: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8\nOutput: 13\nExplanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th smallest number is 13\n\n\n
Example 2:
\n\n\nInput: matrix = [[-5]], k = 1\nOutput: -5\n\n\n
\n
Constraints:
\n\nn == matrix.length == matrix[i].length1 <= n <= 300-109 <= matrix[i][j] <= 109matrix are guaranteed to be sorted in non-decreasing order.1 <= k <= n2\n
Follow up:
\n\nO(1) memory complexity)?O(n) time complexity? The solution may be too advanced for an interview but you may find reading this paper fun.给你一个 n x n 矩阵 matrix ，其中每行和每列元素均按升序排序，找到矩阵中第 k 小的元素。
\n请注意，它是 排序后 的第 k 小元素，而不是第 k 个 不同 的元素。
你必须找到一个内存复杂度优于 O(n2) 的解决方案。
\n\n
示例 1：
\n\n\n输入：matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8\n输出：13\n解释：矩阵中的元素为 [1,5,9,10,11,12,13,13,15]，第 8 小元素是 13\n\n\n
示例 2：
\n\n\n输入：matrix = [[-5]], k = 1\n输出：-5\n\n\n
\n\n
提示：
\n\nn == matrix.lengthn == matrix[i].length1 <= n <= 300-109 <= matrix[i][j] <= 109matrix 中的所有行和列都按 非递减顺序 排列1 <= k <= n2\n\n
进阶：
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