leetcode-problemset/leetcode-cn/originData/maximum-area-of-longest-diagonal-rectangle.json

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            "categoryTitle": "Algorithms",
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            "title": "Maximum Area of Longest Diagonal Rectangle",
            "titleSlug": "maximum-area-of-longest-diagonal-rectangle",
            "content": "<p>You are given a 2D <strong>0-indexed </strong>integer array <code>dimensions</code>.</p>\n\n<p>For all indices <code>i</code>, <code>0 &lt;= i &lt; dimensions.length</code>, <code>dimensions[i][0]</code> represents the length and <code>dimensions[i][1]</code> represents the width of the rectangle<span style=\"font-size: 13.3333px;\"> <code>i</code></span>.</p>\n\n<p>Return <em>the <strong>area</strong> of the rectangle having the <strong>longest</strong> diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the <strong>maximum</strong> area.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> dimensions = [[9,3],[8,6]]\n<strong>Output:</strong> 48\n<strong>Explanation:</strong> \nFor index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) &asymp;<!-- notionvc: 882cf44c-3b17-428e-9c65-9940810216f1 --> 9.487.\nFor index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.\nSo, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> dimensions = [[3,4],[4,3]]\n<strong>Output:</strong> 12\n<strong>Explanation:</strong> Length of diagonal is the same for both which is 5, so maximum area = 12.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= dimensions.length &lt;= 100</code></li>\n\t<li><code><font face=\"monospace\">dimensions[i].length == 2</font></code></li>\n\t<li><code><font face=\"monospace\">1 &lt;= dimensions[i][0], dimensions[i][1] &lt;= 100</font></code></li>\n</ul>\n",
            "translatedTitle": "对角线最长的矩形的面积",
            "translatedContent": "<p>给你一个下标从<strong> 0</strong> 开始的二维整数数组 <code>dimensions</code>。</p>\n\n<p>对于所有下标 <code>i</code>（<code>0 &lt;= i &lt; dimensions.length</code>），<code>dimensions[i][0]</code> 表示矩形 <span style=\"font-size: 13.3333px;\"> <code>i</code></span> 的长度，而 <code>dimensions[i][1]</code> 表示矩形 <span style=\"font-size: 13.3333px;\"> <code>i</code></span> 的宽度。</p>\n\n<p>返回对角线最 <strong>长 </strong>的矩形的<strong> 面积 </strong>。如果存在多个对角线长度相同的矩形，返回面积最<strong> 大 </strong>的矩形的面积。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>dimensions = [[9,3],[8,6]]\n<strong>输出：</strong>48\n<strong>解释：</strong>\n下标 = 0，长度 = 9，宽度 = 3。对角线长度 = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈<!-- notionvc: 882cf44c-3b17-428e-9c65-9940810216f1 --> 9.487。\n下标 = 1，长度 = 8，宽度 = 6。对角线长度 = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10。\n因此，下标为 1 的矩形对角线更长，所以返回面积 = 8 * 6 = 48。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>dimensions = [[3,4],[4,3]]\n<strong>输出：</strong>12\n<strong>解释：</strong>两个矩形的对角线长度相同，为 5，所以最大面积 = 12。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= dimensions.length &lt;= 100</code></li>\n\t<li><code>dimensions[i].length == 2</code></li>\n\t<li><code>1 &lt;= dimensions[i][0], dimensions[i][1] &lt;= 100</code></li>\n</ul>\n",
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