remove-duplicates-from-sorted-array-ii

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Algorithms

Given an integer array <code>nums</code> sorted in non-decreasing order, remove some duplicates <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank">in-place</a> such that each unique element appears at most twice. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part 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.

Return <code>k</code> after placing the final result in the first <code>k</code> slots of <code>nums</code>.

Do not allocate extra space for another array. You must do this by modifying the input array <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank">in-place</a> with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

<pre>
int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i &lt; k; i++) {
 assert nums[i] == expectedNums[i];
}
</pre>

If all assertions pass, then your solution will be accepted.

&nbsp;
Example 1:

<pre>
Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
</pre>

Example 2:

<pre>
Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 104</code></li>
	<li><code>-104 &lt;= nums[i] &lt;= 104</code></li>
	<li><code>nums</code> is sorted in non-decreasing order.</li>
</ul>

removeDuplicates

Remove Duplicates from Sorted Array II

Remove Duplicates from Sorted Array

remove-duplicates-from-sorted-array

删除有序数组中的重复项

Array

Two Pointers

给你一个有序数组 <code>nums</code> ，请你<a href="http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95" target="_blank"> 原地</a> 删除重复出现的元素，使得出现次数超过两次的元素只出现两次 ，返回删除后数组的新长度。

不要使用额外的数组空间，你必须在 <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 并在使用 O(1) 额外空间的条件下完成。

&nbsp;

说明：

为什么返回数值是整数，但输出的答案是数组呢？

请注意，输入数组是以「引用」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。

你可以想象内部操作如下:

<pre>
// nums 是以“引用”方式传递的。也就是说，不对实参做任何拷贝
int len = removeDuplicates(nums);

// 在函数里修改输入数组对于调用者是可见的。
// 根据你的函数返回的长度, 它会打印出数组中 该长度范围内 的所有元素。
for (int i = 0; i &lt; len; i++) {
&nbsp; &nbsp; print(nums[i]);
}
</pre>

&nbsp;

示例 1：

<pre>
输入：nums = [1,1,1,2,2,3]
输出：5, nums = [1,1,2,2,3]
解释：函数应返回新长度 length = <code>5</code>, 并且原数组的前五个元素被修改为 <code>1, 1, 2, 2, 3</code>。 不需要考虑数组中超出新长度后面的元素。
</pre>

示例 2：

<pre>
输入：nums = [0,0,1,1,1,1,2,3,3]
输出：7, nums = [0,0,1,1,2,3,3]
解释：函数应返回新长度 length = <code>7</code>, 并且原数组的前七个元素被修改为&nbsp;<code>0, 0, 1, 1, 2, 3, 3</code>。不需要考虑数组中超出新长度后面的元素。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 104</code></li>
	<li><code>-104 &lt;= nums[i] &lt;= 104</code></li>
	<li><code>nums</code> 已按升序排列</li>
</ul>