leetcode 80 Remove Duplicates from Sorted Array II

题目

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.

解法

class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        j = 1
        for i in range(1, len(nums)):
            if j == 1 or nums[i] != nums[j - 2]:
                nums[j] = nums[i]
                j += 1
        return j
                

显然和前面的merge lists (26) 还有remove duplicates (88) 是一种类型. 从前面只留单个不重复得元素还看不真切，现在看就是快慢指针。其中快指针i遍历List, 然后慢指针只在遍历到的元素满足条件后，把满足条件的元素填入当前位置，然后向前一步。
什么条件呢： 当前元素和慢指针向前两位的数不同。因为是有序list所以不用考虑前一位。

最终逻辑

1. 慢指针指的是list当前需要确定值得位置，
2. 那么值就是不停由快指针向前方找，
3. 找到那个满足List已经确定的值最后两位不一样的为止，
4. 填入List,
   t已经确定的值最后两位不一样的为止，
5. 填入List,
6. 更新慢指针。