题目：Follow up for "Remove Duplicates":

What if duplicates are allowed at most twice?

For example,

Given sorted array 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. It doesn't matter what you leave beyond the new length.

解答：

class Solution {

public:

    int removeDuplicates(vector<int>& nums) {

        int size = nums.size();

        if(size <= 2)

            return size;

        int cur_index = 0;

        int p = 1;

        int q = 2;

        while(q < size)

        {

            if(nums[cur_index] != nums[p])

            {

                nums[++cur_index] = nums[p++];

                q = p + 1;

            }

            else

            {

                if(nums[p] != nums[q])

                {

                    nums[++cur_index] = nums[p++];

                    q = p + 1;

                }

                else

                {

                    p++;

                    q = p + 1;

                }

            }

        }

        nums[++cur_index] = nums[p++];

        return cur_index + 1;      

    }

};

 

其中，cur_index一直控制着最终应该返回的只允许两次重复的边界，而p是当前正在判断的数字，q为其下一个数字。

自己的题解过程，看起来很复杂，其实，仔细一想，只需要一个变量记录当前的位置，需要一个变量来迭代的往后走，判断该变量指向的位置的前两个位置是否与迭代变量所指向的相同。

优秀答案：

class Solution {

public:

    int removeDuplicates(vector<int>& nums) {

        int size = nums.size();

        if(size <= 2)

            return size;

        int index = 2;

        for(int i = 2;i < size;i++)

        {

            if(nums[index - 2] != nums[i])

            {

                nums[index++] = nums[i];

            }

        }

        return index;

    }

};