# 【LeetCode】 26. Remove Duplicates from Sorted Array

## Description
> 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.

> 給一個已經排序完的陣列nums，用in-place演算法移除重複的元素，使每個元素都只會出現一次，並回傳新的長度。
> 不要去分配額外的空間給其他陣列，你應該直接修改輸入的陣列，空間複雜度應為O(1)。

## Example:

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.

## Solution

* 紀錄第一個元素在`temp`。
* 遍歷陣列，如果和`temp`一樣，就移除；否則將`temp`更新。
* 其他注意：用`nums.size()`的速度很慢，你可以另外用一個空間去儲存、操作它。
* 其他注意：用這個Code跑LeetCode的速度會輸給超多人，原因是其他人沒有使用in-place演算法，但評測程式不會抓。

### Code

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int s = nums.size();
        if(s==0)return 0;
        int temp = nums[0];
        for(int i =1;i<s;)
        {
            if(temp == nums[i])
            {
                s--;
                nums.erase(nums.begin() + i);
            }
            else
            {
                temp = nums[i];
                i++;
            }
        }
        return s;
    }
};

###### tags: `LeetCode` `C++`