leetcode-problemset/leetcode-cn/originData/remove-duplicates-from-sorted-array-ii.json

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            "title": "Remove Duplicates from Sorted Array II",
            "titleSlug": "remove-duplicates-from-sorted-array-ii",
            "content": "<p>Given an integer array <code>nums</code> sorted in <strong>non-decreasing order</strong>, remove some duplicates <a href=\"https://en.wikipedia.org/wiki/In-place_algorithm\" target=\"_blank\"><strong>in-place</strong></a> such that each unique element appears <strong>at most twice</strong>. The <strong>relative order</strong> of the elements should be kept the <strong>same</strong>.</p>\n\n<p>Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the <strong>first part</strong> of the array <code>nums</code>. More formally, if there are <code>k</code> elements after removing the duplicates, then the first <code>k</code> elements of <code>nums</code>&nbsp;should hold the final result. It does not matter what you leave beyond the first&nbsp;<code>k</code>&nbsp;elements.</p>\n\n<p>Return <code>k</code><em> after placing the final result in the first </em><code>k</code><em> slots of </em><code>nums</code>.</p>\n\n<p>Do <strong>not</strong> allocate extra space for another array. You must do this by <strong>modifying the input array <a href=\"https://en.wikipedia.org/wiki/In-place_algorithm\" target=\"_blank\">in-place</a></strong> with O(1) extra memory.</p>\n\n<p><strong>Custom Judge:</strong></p>\n\n<p>The judge will test your solution with the following code:</p>\n\n<pre>\nint[] nums = [...]; // Input array\nint[] expectedNums = [...]; // The expected answer with correct length\n\nint k = removeDuplicates(nums); // Calls your implementation\n\nassert k == expectedNums.length;\nfor (int i = 0; i &lt; k; i++) {\n    assert nums[i] == expectedNums[i];\n}\n</pre>\n\n<p>If all assertions pass, then your solution will be <strong>accepted</strong>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,1,1,2,2,3]\n<strong>Output:</strong> 5, nums = [1,1,2,2,3,_]\n<strong>Explanation:</strong> Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.\nIt does not matter what you leave beyond the returned k (hence they are underscores).\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,0,1,1,1,1,2,3,3]\n<strong>Output:</strong> 7, nums = [0,0,1,1,2,3,3,_,_]\n<strong>Explanation:</strong> Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.\nIt does not matter what you leave beyond the returned k (hence they are underscores).\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>nums</code> is sorted in <strong>non-decreasing</strong> order.</li>\n</ul>\n",
            "translatedTitle": "删除有序数组中的重复项 II",
            "translatedContent": "<p>给你一个有序数组 <code>nums</code> ，请你<strong><a href=\"http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95\" target=\"_blank\"> 原地</a></strong> 删除重复出现的元素，使得出现次数超过两次的元素<strong>只出现两次</strong> ，返回删除后数组的新长度。</p>\n\n<p>不要使用额外的数组空间，你必须在 <strong><a href=\"https://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95\" target=\"_blank\">原地 </a>修改输入数组 </strong>并在使用 O(1) 额外空间的条件下完成。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>说明：</strong></p>\n\n<p>为什么返回数值是整数，但输出的答案是数组呢？</p>\n\n<p>请注意，输入数组是以<strong>「引用」</strong>方式传递的，这意味着在函数里修改输入数组对于调用者是可见的。</p>\n\n<p>你可以想象内部操作如下:</p>\n\n<pre>\n// <strong>nums</strong> 是以“引用”方式传递的。也就是说，不对实参做任何拷贝\nint len = removeDuplicates(nums);\n\n// 在函数里修改输入数组对于调用者是可见的。\n// 根据你的函数返回的长度, 它会打印出数组中<strong> 该长度范围内</strong> 的所有元素。\nfor (int i = 0; i &lt; len; i++) {\n&nbsp; &nbsp; print(nums[i]);\n}\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,1,1,2,2,3]\n<strong>输出：</strong>5, nums = [1,1,2,2,3]\n<strong>解释：</strong>函数应返回新长度 length = <strong><code>5</code></strong>, 并且原数组的前五个元素被修改为 <strong><code>1, 1, 2, 2, 3</code></strong>。 不需要考虑数组中超出新长度后面的元素。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [0,0,1,1,1,1,2,3,3]\n<strong>输出：</strong>7, nums = [0,0,1,1,2,3,3]\n<strong>解释：</strong>函数应返回新长度 length = <strong><code>7</code></strong>, 并且原数组的前五个元素被修改为&nbsp;<strong><code>0, 0, 1, 1, 2, 3, 3</code></strong>。不需要考虑数组中超出新长度后面的元素。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>nums</code> 已按升序排列</li>\n</ul>\n",
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