Description:
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
Example:
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.
Link:
https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/description/
解题方法：
多加一个变量记录当前数字出现的次数，其余做法与Remove Duplicates from Sorted Array的做法一样
Time Complexity：
O(N)
完整代码：
int removeDuplicates(vector<int>& nums)
{
if(!nums.size())
return 0;
int time = 1, position = 1;
for(int i = 1; i < nums.size(); i++)
{
if(nums[i] == nums[i - 1])
{
if(time < 2)
{
nums[position] = nums[i];
position++;
}
time++;
}
else
{
time = 1;
nums[position] = nums[i];
position++;
}
}
return position;
}
