Approach

# 出現沒有記錄過的值就加進陣列，回傳值剛好是陣列大小 + 1

Time Complexity

O(n)

Space Complexity

O(1)

Code

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int n = 0;

        for (int i = 1 ; i < nums.size() ; i++) {
            if (nums[i] != nums[n]) {
                nums[++n] = nums[i];
            }
        }

        return n + 1;
    }
};