Question
Formatted question description: https://leetcode.ca/all/80.html
80. Remove Duplicates from Sorted Array II
Given a sorted array nums, remove the duplicates inplace such that duplicates appeared at most twice and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array inplace with O(1) extra memory.
Example 1:
Given nums = [1,1,1,2,2,3],
Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,0,1,1,1,1,2,3,3],
Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.
It doesn't matter what values are set beyond the returned length.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}
@tagarray
Algorithm
The maximum number of repetitions allowed is two, then a variable cnt can be used to record and several repetitions are allowed, cnt is initialized to 1, if there is one repetition, cnt is decremented by 1, then the next time there is a repetition, the fast pointer is directly moving forward, if it is not repeated at this time, cnt is restored to 1.
Since the entire array is ordered, once a nonrepeating number appears, it must be greater than this number, and there will be no more duplicates after this number.
Code
Java

public class Remove_Duplicates_from_Sorted_Array_II { class Solution { public int removeDuplicates(int[] nums) { int i = 0; for (int num : nums) { if (i < 2  num > nums[i  2]) { nums[i++] = num; } } return i; } } public int removeDuplicates(int[] nums) { if (nums.length == 0) { return 0; } int pre = 0, cur = 1, dupAllowed = 1, n = nums.length; while (cur < n) { if (nums[pre] == nums[cur] && dupAllowed == 0) { ++cur; } else { if (nums[pre] == nums[cur]) { dupAllowed; } else { dupAllowed = 1; } nums[++pre] = nums[cur++]; } } return pre + 1; } }

// OJ: https://leetcode.com/problems/removeduplicatesfromsortedarrayii/ // Time: O(N) // Space: O(1) class Solution { public: int removeDuplicates(vector<int>& A) { int j = 0; for (int i = 0; i < A.size(); ++i) { if (j  2 < 0  A[j  2] != A[i]) A[j++] = A[i]; } return j; } };

class Solution(object): def removeDuplicates(self, nums): """ :type nums: List[int] :rtype: int """ if len(nums) <= 2: return len(nums) cnt = 0 j = 1 for i in range(1, len(nums)): if nums[i] == nums[i  1]: cnt += 1 if cnt < 2: nums[j] = nums[i] j += 1 else: nums[j] = nums[i] j += 1 cnt = 0 return j