leetcode-problemset/leetcode-cn/originData/median-of-two-sorted-arrays.json

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            "title": "Median of Two Sorted Arrays",
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            "content": "<p>Given two sorted arrays <code>nums1</code> and <code>nums2</code> of size <code>m</code> and <code>n</code> respectively, return <strong>the median</strong> of the two sorted arrays.</p>\n\n<p>The overall run time complexity should be <code>O(log (m+n))</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,3], nums2 = [2]\n<strong>Output:</strong> 2.00000\n<strong>Explanation:</strong> merged array = [1,2,3] and median is 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,2], nums2 = [3,4]\n<strong>Output:</strong> 2.50000\n<strong>Explanation:</strong> merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>nums1.length == m</code></li>\n\t<li><code>nums2.length == n</code></li>\n\t<li><code>0 &lt;= m &lt;= 1000</code></li>\n\t<li><code>0 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= m + n &lt;= 2000</code></li>\n\t<li><code>-10<sup>6</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
            "translatedTitle": "寻找两个正序数组的中位数",
            "translatedContent": "<p>给定两个大小分别为 <code>m</code> 和 <code>n</code> 的正序（从小到大）数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>。请你找出并返回这两个正序数组的 <strong>中位数</strong> 。</p>\n\n<p>算法的时间复杂度应该为 <code>O(log (m+n))</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [1,3], nums2 = [2]\n<strong>输出：</strong>2.00000\n<strong>解释：</strong>合并数组 = [1,2,3] ，中位数 2\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [1,2], nums2 = [3,4]\n<strong>输出：</strong>2.50000\n<strong>解释：</strong>合并数组 = [1,2,3,4] ，中位数 (2 + 3) / 2 = 2.5\n</pre>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>nums1.length == m</code></li>\n\t<li><code>nums2.length == n</code></li>\n\t<li><code>0 &lt;= m &lt;= 1000</code></li>\n\t<li><code>0 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= m + n &lt;= 2000</code></li>\n\t<li><code>-10<sup>6</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
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