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57 lines
2.0 KiB
Markdown
57 lines
2.0 KiB
Markdown
# [80. Remove Duplicates from Sorted Array II](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/)
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## 题目
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Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
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Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
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Example 1:
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```c
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Given nums = [1,1,1,2,2,3],
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Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
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It doesn't matter what you leave beyond the returned length.
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```
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Example 2:
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```c
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Given nums = [0,0,1,1,1,1,2,3,3],
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Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.
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It doesn't matter what values are set beyond the returned length.
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```
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Clarification:
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Confused why the returned value is an integer but your answer is an array?
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Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
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Internally you can think of this:
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```c
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// nums is passed in by reference. (i.e., without making a copy)
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int len = removeElement(nums, val);
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// any modification to nums in your function would be known by the caller.
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// using the length returned by your function, it prints the first len elements.
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for (int i = 0; i < len; i++) {
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print(nums[i]);
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}
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```
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## 题目大意
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给定一个有序数组 nums，对数组中的元素进行去重，使得原数组中的每个元素最多暴露 2 个。最后返回去重以后数组的长度值。
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## 解题思路
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- 问题提示有序数组，一般最容易想到使用双指针的解法，双指针的关键点：移动两个指针的条件。
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- 在该题中移动的条件：快指针从头遍历数组，慢指针指向修改后的数组的末端，当慢指针指向倒数第二个数与快指针指向的数不相等时，才移动慢指针，同时赋值慢指针。
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- 处理边界条件：当数组小于两个元素时，不做处理。
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