Given a sorted array nums, remove the duplicates in-place 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 in-place with O(1) extra memory.

Example 1:

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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:

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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:

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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]);
}

分析

跟26. Remove Duplicates from Sorted Array比，可以重复一次，代码结构是一样的，只不过需要跳过一个重复记录。

代码

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class  {
public:
    int removeDuplicates(vector<int>& nums) {
        int len = nums.length();
        if (len <=2) return len;

        int index = 2;
        for (int i = 2; i < len; ++i)
        {
            if (nums[i] != nums[index-2]) {
                nums[index++] = nums[i];
            }
        }
        return index;
    }
};