题目描述：

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：遍历数组，用变量count记录重复多余两次的元素出现的次数，将每个元素向前移动count个单位。

代码：

int Solution::removeDuplicates(int A[], int n)
{
    int count = 0;
    bool first = true;
    for(int i = 0;i < n;i++)
    {
        if(i != 0)
        {
            if(A[i] == A[i-1] && first)
            {
                first = false;
                A[i-count] = A[i];
                continue;
            }
            else if(A[i] == A[i-1] && !first)
            {
                count++;
                A[i-count] = A[i];
                continue;
            }
            else
            {
                first = true;
                A[i-count] = A[i];
                continue;
            }
        }
    }
    return n - count;
}

LeetCode:Remove Duplicates from Sorted Array II