# 80. Remove Duplicates from Sorted Array II

## problem description

Given a sorted array _nums_, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that duplicates appeared at most _twice_ and return the new length.

Do not allocate extra space for another array, you must do this by **modifying the input array** [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) with O(1) extra memory.

**Example 1:**

```
Given nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.

It doesn't matter what you leave beyond the returned length.
```

**Example 2:**

```
Given nums = [0,0,1,1,1,1,2,3,3],

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.

It doesn't matter what values are set beyond the returned length.
```

**Clarification:**

Confused why the returned value is an integer but your answer is an array?

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.

Internally you can think of this:

```
// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}
```

## algorithm thought

这题目其实很简单，怎么样都能做得出。只是有些人做的简单有些人做的复杂。首先用for range语句，得到nums所有的数字。然后和nums\[pos-2]相比，如果大于的话，那么当前肯定要加，但是如果等于的话，那肯定就是前面已经放置了两个相同的数，就可以不用再写了。

## code

```cpp
class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int pos=0;
        for(auto num:nums){
            if(pos<2||num>nums[pos-2]){
                nums[pos++]=num;
            }
        }
        return pos;
    }
};
```

## algorithm analysis

一次遍历出结果，时间复杂度是O(n)
