Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums. Return k.

給定一個排序數組，你需要在原地刪除重複出現的元素，使得每個元素只出現一次，返回移除後數組的新長度。

不要使用額外的數組空間，你必須在原地修改輸入數組並使用O（1）額外空間的條件下完成。

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

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 k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

<解題>

1. 條件檢查：程式碼確保nums不為空&&長度不為0
2. 設定初始值=1和不重複元素值為目前第一個元素
3. 將非重複元素放入答案的位置，指標繼續往後動

class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int answer = 1; // 初始化答案為1，因為第一個元素肯定是不重複的
        int prev = nums[0]; // 用來記錄前一個非重複元素的值
        
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != prev) {
                nums[answer] = nums[i]; // 將非重複元素放入答案的位置
                prev = nums[i]; // 更新prev的值為目前的元素
                answer++; // 答案位置往後移動
            }
        }
        
        return answer;
    }
}