###### tags: `Leetcode` `medium` `pointer` `python`

# 80. Remove Duplicates from Sorted Array II

## [題目連結:]https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/

## 題目:
Given an integer array ```nums``` sorted in **non-decreasing order**, remove some duplicates ```in-place``` such that each unique element appears at **most twice**. The **relative order** of the elements should be kept the **same**.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the **first part** of the array ```nums```. More formally, if there are ```k``` elements after removing the duplicates, then the first ```k``` elements of nums should hold the final result. It does not matter what you leave beyond the first ```k``` elements.

Return ```k``` after placing the final result in the first ```k``` slots of ```nums```.

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

**Custom Judge:**

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be **accepted**.

**Example 1:**

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

**Example 2:**

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

## 解題想法:
* 題目為給一非遞減的數組
* 要求每個數字最多出現兩次
* 相同的數字必須相鄰
* return 刪除多餘的數字後的長度
* modify in-place O(1) extra memory.
* 往pointer思考
* 想法:
* 設2個pointer:
* fast=0
* 正常遍歷所有數字
* slow=0
* 指向多餘的第一個位置
* 當nums[fast]遍歷到新的數
* 此新數與nums[slow-2]
* 不相同時，表示該新字數不為重複，
* 將其取代nums[slow]
* fast、slow皆向後移動
* 否則，continue讓fast繼續正常移動

* trace: nums=[1,1,1,2,2,3]

ans: nums[fast]==nums[slow-2] : slow停
            f
    1   1   1   2   2   3
            s
ans: nums[fast]!=nums[slow-2]: 將nums[fast]取代nums[slow] 兩pointer均移動 (2取代1)
                  f
   '1'   1   1   '2'   2   3
             s
ans: nums[fast]!=nums[slow-2]: 將nums[fast]取代nums[slow] 兩pointer均移動 (2取代2 沒差>< )
                     f
    1  '1'   2   2   2   3
                 s
ans: nums[fast]!=nums[slow-2]: 將nums[fast]取代nums[slow] 兩pointer均移動 (3取代2)
                         f
    1   1  '2'   2   2   3
                     s
final: 
                            f
    1   1   2   2   3   3
                        s (剛好slow所在pos為len(刪除重複後的nums)->[1,1,2,2,3])

## Python:

class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums)<3:
            return len(nums)
        #前面兩個位置(pos:0,1)的數正常保留，從2開始
        fast=2
        slow=2
        for fast in range(2,len(nums)):
            if nums[fast]==nums[slow-2]:
                continue
            else:
                nums[slow]=nums[fast]
                slow+=1
        return slow

result = Solution()
nums = [1,1,1,2,2,3]
ans = result.removeDuplicates(nums)
print(ans,nums)