Given a sorted array *nums*, remove the duplicates **in-place** such that each element appear only *once* and return the new length.

Do not allocate extra space for another array, you must do this by **modifying the input array in-place** with O(1) extra memory.

**Example 1:**

```
Given nums = [1,1,2],
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:**

```
Given nums = [0,0,1,1,1,2,2,3,3,4],
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.
```

**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]);
}
```

**Solution:**

```
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
nums.erase(unique(nums.begin(),nums.end()),nums.end());
return nums.size();
}
};
```

Standard way:

```
int count = 0;
for(int i = 1; i < n; i++){
if(A[i] == A[i-1]) count++;
else A[i-count] = A[i];
}
return n-count;
```