leetcode 之Remove Duplicates from Sorted Array（2）

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]

    之前的想法是再加一个计数的变量就行了

    

int removeDeplicates1(int A[], int n)
{
    int index = 0, count = 1;
    for (int i = 1; i < n; i++)
    {
        if (A[index] != A[i])
        {
            A[index++] = A[i];
            count = 1;
        }
        else 
        {
            if (count <= 2)
            {
                A[index] = A[i];
                count++;
            }
        }
    }

    return index + 1;
}

       后来看到了一个更简洁的做法，这个想法很巧妙啊：因为是排好序的，所以只需拿第三个和第一个比，不同的话保存即可。

int removeDeplicates1(int A[], int n)
{
    int index = 2;
    for (int i = 2; i < n; i++)
    {
        if (A[index-2] != A[i])
        {
            A[index++] = A[i];
        }
        
    }

    return index;
}