Leetcode题目：Remove Duplicates from Sorted Array II

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;
    }
};