Remove Duplicates from Sorted Array II--LeetCode

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3].

思路：需要一个标志位记录当前这个数出现的次数，如果和前面的不相等，肯定得覆盖，并且记录出现了两次，如果和前面相等，那么出现次数多一次，如果出现的次数已经为两次，那么也允许覆盖。

#include <iostream>
#include <vector>
using namespace std;
/*
 Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3].
*/ 

int RemoveDuplicate(vector<int>& vec)
{
	if(vec.size()<=2)
		return vec.size();
	int i,flag=1,j;
	i =0;
	for(j=1;j<vec.size();j++)
	{
		if(vec[j] != vec[j-1])
		{
			i++;
			vec[i] = vec[j];
			flag =1;
		}
		else
		{
			flag++;
			if(flag==2)
			{
				i++;
				vec[i] = vec[j];
			}
			
		}
	}
	return i+1;
}

int main()
{
	int array[]={1,1,2,2,3,4,4,5,6,7};
	vector<int> vec(array,array+sizeof(array)/sizeof(int));
	cout<<RemoveDuplicate(vec);
	return 0;
}

同样使用一个标志位来表示

int removeDuplicates(vector<int>& vec)
{
	int i=0,j=1,flag=0;
	for(j=1;j<vec.size();j++)
	{
	  if(vec[j] == vec[i] && !flag)
	  {
	  	i++;
	  	vec[i] = vec[j];
	  	flag =1;
	  }
	  if(vec[j]!=vec[i])
	  {
	  	i++;
	  	vec[i] = vec[j];
	  	flag =0;
	  }
	}
	return i+1;
}