leetcode-problemset/leetcode-cn/originData/count-negative-numbers-in-a-sorted-matrix.json

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        "question": {
            "questionId": "1476",
            "questionFrontendId": "1351",
            "categoryTitle": "Algorithms",
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            "title": "Count Negative Numbers in a Sorted Matrix",
            "titleSlug": "count-negative-numbers-in-a-sorted-matrix",
            "content": "<p>Given a <code>m x n</code> matrix <code>grid</code> which is sorted in non-increasing order both row-wise and column-wise, return <em>the number of <strong>negative</strong> numbers in</em> <code>grid</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]\n<strong>Output:</strong> 8\n<strong>Explanation:</strong> There are 8 negatives number in the matrix.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> grid = [[3,2],[1,0]]\n<strong>Output:</strong> 0\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == grid.length</code></li>\n\t<li><code>n == grid[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 100</code></li>\n\t<li><code>-100 &lt;= grid[i][j] &lt;= 100</code></li>\n</ul>\n\n<p>&nbsp;</p>\n<strong>Follow up:</strong> Could you find an <code>O(n + m)</code> solution?",
            "translatedTitle": "统计有序矩阵中的负数",
            "translatedContent": "<p>给你一个&nbsp;<code>m&nbsp;* n</code>&nbsp;的矩阵&nbsp;<code>grid</code>，矩阵中的元素无论是按行还是按列，都以非递增顺序排列。&nbsp;请你统计并返回&nbsp;<code>grid</code>&nbsp;中 <strong>负数</strong> 的数目。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]\n<strong>输出：</strong>8\n<strong>解释：</strong>矩阵中共有 8 个负数。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>grid = [[3,2],[1,0]]\n<strong>输出：</strong>0\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>m == grid.length</code></li>\n\t<li><code>n == grid[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 100</code></li>\n\t<li><code>-100 &lt;= grid[i][j] &lt;= 100</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你可以设计一个时间复杂度为 <code>O(n + m)</code> 的解决方案吗？</p>\n",
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