leetcode-problemset/leetcode-cn/originData/rotting-oranges.json

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            "title": "Rotting Oranges",
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            "content": "<p>You are given an <code>m x n</code> <code>grid</code> where each cell can have one of three values:</p>\n\n<ul>\n\t<li><code>0</code> representing an empty cell,</li>\n\t<li><code>1</code> representing a fresh orange, or</li>\n\t<li><code>2</code> representing a rotten orange.</li>\n</ul>\n\n<p>Every minute, any fresh orange that is <strong>4-directionally adjacent</strong> to a rotten orange becomes rotten.</p>\n\n<p>Return <em>the minimum number of minutes that must elapse until no cell has a fresh orange</em>. If <em>this is impossible, return</em> <code>-1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/02/16/oranges.png\" style=\"width: 650px; height: 137px;\" />\n<pre>\n<strong>Input:</strong> grid = [[2,1,1],[1,1,0],[0,1,1]]\n<strong>Output:</strong> 4\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> grid = [[2,1,1],[0,1,1],[1,0,1]]\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> grid = [[0,2]]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Since there are already no fresh oranges at minute 0, the answer is just 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;= 10</code></li>\n\t<li><code>grid[i][j]</code> is <code>0</code>, <code>1</code>, or <code>2</code>.</li>\n</ul>\n",
            "translatedTitle": "腐烂的橘子",
            "translatedContent": "<p>在给定的&nbsp;<code>m x n</code>&nbsp;网格<meta charset=\"UTF-8\" />&nbsp;<code>grid</code>&nbsp;中，每个单元格可以有以下三个值之一：</p>\n\n<ul>\n\t<li>值&nbsp;<code>0</code>&nbsp;代表空单元格；</li>\n\t<li>值&nbsp;<code>1</code>&nbsp;代表新鲜橘子；</li>\n\t<li>值&nbsp;<code>2</code>&nbsp;代表腐烂的橘子。</li>\n</ul>\n\n<p>每分钟，腐烂的橘子&nbsp;<strong>周围&nbsp;4 个方向上相邻</strong> 的新鲜橘子都会腐烂。</p>\n\n<p>返回 <em>直到单元格中没有新鲜橘子为止所必须经过的最小分钟数。如果不可能，返回&nbsp;<code>-1</code></em>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/02/16/oranges.png\" style=\"height: 137px; width: 650px;\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>grid = [[2,1,1],[1,1,0],[0,1,1]]\n<strong>输出：</strong>4\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>grid = [[2,1,1],[0,1,1],[1,0,1]]\n<strong>输出：</strong>-1\n<strong>解释：</strong>左下角的橘子（第 2 行， 第 0 列）永远不会腐烂，因为腐烂只会发生在 4 个正向上。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>grid = [[0,2]]\n<strong>输出：</strong>0\n<strong>解释：</strong>因为 0 分钟时已经没有新鲜橘子了，所以答案就是 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;= 10</code></li>\n\t<li><code>grid[i][j]</code> 仅为&nbsp;<code>0</code>、<code>1</code>&nbsp;或&nbsp;<code>2</code></li>\n</ul>\n",
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