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;
}
};