leetcode-problemset/leetcode-cn/originData/missing-number.json

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            "title": "Missing Number",
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            "content": "<p>Given an array <code>nums</code> containing <code>n</code> distinct numbers in the range <code>[0, n]</code>, return <em>the only number in the range that is missing from the array.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,0,1]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,1]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [9,6,4,2,3,5,7,0,1]\n<strong>Output:</strong> 8\n<strong>Explanation:</strong> n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= n</code></li>\n\t<li>All the numbers of <code>nums</code> are <strong>unique</strong>.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up:</strong> Could you implement a solution using only <code>O(1)</code> extra space complexity and <code>O(n)</code> runtime complexity?</p>\n",
            "translatedTitle": "丢失的数字",
            "translatedContent": "<p>给定一个包含 <code>[0, n]</code>&nbsp;中&nbsp;<code>n</code>&nbsp;个数的数组 <code>nums</code> ，找出 <code>[0, n]</code> 这个范围内没有出现在数组中的那个数。</p>\n\n<ul>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3,0,1]\n<strong>输出：</strong>2\n<b>解释：</b>n = 3，因为有 3 个数字，所以所有的数字都在范围 [0,3] 内。2 是丢失的数字，因为它没有出现在 nums 中。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [0,1]\n<strong>输出：</strong>2\n<b>解释：</b>n = 2，因为有 2 个数字，所以所有的数字都在范围 [0,2] 内。2 是丢失的数字，因为它没有出现在 nums 中。</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [9,6,4,2,3,5,7,0,1]\n<strong>输出：</strong>8\n<b>解释：</b>n = 9，因为有 9 个数字，所以所有的数字都在范围 [0,9] 内。8 是丢失的数字，因为它没有出现在 nums 中。</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [0]\n<strong>输出：</strong>1\n<b>解释：</b>n = 1，因为有 1 个数字，所以所有的数字都在范围 [0,1] 内。1 是丢失的数字，因为它没有出现在 nums 中。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= n</code></li>\n\t<li><code>nums</code> 中的所有数字都 <strong>独一无二</strong></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你能否实现线性时间复杂度、仅使用额外常数空间的算法解决此问题?</p>\n",
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