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"categoryTitle": "Algorithms",
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"title": "Kth Smallest Element in a Sorted Matrix",
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"content": "<p>Given an <code>n x n</code> <code>matrix</code> where each of the rows and columns is sorted in ascending order, return <em>the</em> <code>k<sup>th</sup></code> <em>smallest element in the matrix</em>.</p>\n\n<p>Note that it is the <code>k<sup>th</sup></code> smallest element <strong>in the sorted order</strong>, not the <code>k<sup>th</sup></code> <strong>distinct</strong> element.</p>\n\n<p>You must find a solution with a memory complexity better than <code>O(n<sup>2</sup>)</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8\n<strong>Output:</strong> 13\n<strong>Explanation:</strong> The elements in the matrix are [1,5,9,10,11,12,13,<u><strong>13</strong></u>,15], and the 8<sup>th</sup> smallest number is 13\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> matrix = [[-5]], k = 1\n<strong>Output:</strong> -5\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == matrix.length == matrix[i].length</code></li>\n\t<li><code>1 &lt;= n &lt;= 300</code></li>\n\t<li><code>-10<sup>9</sup> &lt;= matrix[i][j] &lt;= 10<sup>9</sup></code></li>\n\t<li>All the rows and columns of <code>matrix</code> are <strong>guaranteed</strong> to be sorted in <strong>non-decreasing order</strong>.</li>\n\t<li><code>1 &lt;= k &lt;= n<sup>2</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up:</strong></p>\n\n<ul>\n\t<li>Could you solve the problem with a constant memory (i.e., <code>O(1)</code> memory complexity)?</li>\n\t<li>Could you solve the problem in <code>O(n)</code> time complexity? The solution may be too advanced for an interview but you may find reading <a href=\"http://www.cse.yorku.ca/~andy/pubs/X+Y.pdf\" target=\"_blank\">this paper</a> fun.</li>\n</ul>\n",
"translatedTitle": "有序矩阵中第 K 小的元素",
"translatedContent": "<p>给你一个&nbsp;<code>n x n</code><em>&nbsp;</em>矩阵&nbsp;<code>matrix</code> ,其中每行和每列元素均按升序排序,找到矩阵中第 <code>k</code> 小的元素。<br />\n请注意它是 <strong>排序后</strong> 的第 <code>k</code> 小元素,而不是第 <code>k</code> 个 <strong>不同</strong> 的元素。</p>\n\n<p>你必须找到一个内存复杂度优于&nbsp;<code>O(n<sup>2</sup>)</code> 的解决方案。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8\n<strong>输出:</strong>13\n<strong>解释:</strong>矩阵中的元素为 [1,5,9,10,11,12,13,<strong>13</strong>,15],第 8 小元素是 13\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>matrix = [[-5]], k = 1\n<strong>输出:</strong>-5\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>n == matrix.length</code></li>\n\t<li><code>n == matrix[i].length</code></li>\n\t<li><code>1 &lt;= n &lt;= 300</code></li>\n\t<li><code>-10<sup>9</sup> &lt;= matrix[i][j] &lt;= 10<sup>9</sup></code></li>\n\t<li>题目数据 <strong>保证</strong> <code>matrix</code> 中的所有行和列都按 <strong>非递减顺序</strong> 排列</li>\n\t<li><code>1 &lt;= k &lt;= n<sup>2</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶:</strong></p>\n\n<ul>\n\t<li>你能否用一个恒定的内存(即 <code>O(1)</code> 内存复杂度)来解决这个问题?</li>\n\t<li>你能在 <code>O(n)</code> 的时间复杂度下解决这个问题吗?这个方法对于面试来说可能太超前了,但是你会发现阅读这篇文章(&nbsp;<a href=\"http://www.cse.yorku.ca/~andy/pubs/X+Y.pdf\" target=\"_blank\">this paper</a>&nbsp;)很有趣。</li>\n</ul>\n",
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