# LeetCode 26. Remove Duplicates from Sorted Array



Given a sorted array _nums_, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each element appears only _once_ and returns 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.

**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 a 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]);
}
```

**Example 1:**

```
Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.
```

**Example 2:**

```
Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4]
Explanation: Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length.
```

**Constraints:**

* `0 <= nums.length <= 3 * 10^4`
* `-10^4 <= nums[i] <= 10^4`
* `nums` is sorted in ascending order.

## Solution:

[English Version in Youtube](https://youtu.be/ifVpYbnm05A)

[中文版解答Youtube Link](https://youtu.be/Dy9uPt-HvYQ)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1Gy4y1a7AB/)

```
class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int pos = 0;
        for (int n : nums) {
            if (pos == 0 || n > nums[pos-1]) {
                nums[pos] = n;
                pos++;
            }
        }
        return pos;
    }
};
```
