嗯 这题没有想好，这个做法是看的网络上的，然后复现了一下

class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        idx = 0
        count = 0
        for i in range(len(nums)):
            if i != 0 and nums[i] == nums[i-1]:
                count += 1
                if count >= 3:
                    continue
            else:
                count = 1
            nums[idx] = nums[i]
            idx += 1
        return idx
                

另一种做法是用一个指针J 来指向当前位置和下一个位置，判断是否相等，如果相等 两个数字都存下来，如果不同，存下来当前的数字，还要注意去重

class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        
        n = len(nums)
        if n == 0 or n == 1 or n == 2:
            return n
        i = 0
        j = 0
        while i < n and j < n:
            if j + 1 < n and nums[j] == nums[j+1]:
                nums[i] = nums[j]
                nums[i+1] = nums[j+1]
                i += 2
                while j < n and nums[j] == nums[i - 1]:
                    j += 1
            else:
                nums[i] = nums[j]
                i += 1
                j += 1
        return i