Remove Duplicates from Sorted ListII

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].

原始思路:

找一个标签数字，遍历向量元素，如果相等，则加1， 如果大于2,则continue遍历，如果不同 标签数字换成新的数字

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
	vector<int> a = { 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4,4 };
	int k = 0;
	int key = a[0];
	int index = 0;
	int i = 0;
		while (i < a.size())
		{
			if (a[i] == key)
			{
				k++;
				if (k<3)
			     a[index++] = key;
				else 
				{
					i++;
					continue;
					
				}
			}
			else
			{
				key = a[i];
				a[index++] = key;
				k = 1;
			}
			
			i++;
		}
		
	

	for (int i = 0; i < a.size(); i++)
		cout << a[i] << endl;

	system("pause");
	return 0;
}

应为允许一个重复，所以判断就不是相邻而是相隔一个的数字是否不一样。

不一样就需要更新值，但注意到这题更新数组不能是当前的状态，而需要是更新前一个不同的数，用temp存当前值以便于下一次赋值。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
	vector<int> a = { 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4,4 };
	int k = 0;
	/*int key = a[0];
	int index = 0;
	int i = 0;
		while (i < a.size())
		{
			if (a[i] == key)
			{
				k++;
				if (k<3)
			     a[index++] = key;
				else 
				{
					i++;
					continue;
					
				}
			}
			else
			{
				key = a[i];
				a[index++] = key;
				k = 1;
			}
			
			i++;
		}
		
	*/
	int num = 1, i, temp = a[1];
	for (i = 2; i<a.size(); ++i)
	if (a[i] != a[i - 2])
	{
		a[num++] = temp;
		temp = a[i];
	}
	a[num++] = temp;

	for (int i = 0; i < a.size(); i++)
		cout << a[i] << endl;

	system("pause");
	return 0;
}