Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array A = [1,1,2],

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

这题与Remove Duplicates from Sorted List类似，只不过换成了数组，要把排好序的数组的元素变为不重复，且要求原地操作。

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        int new_arr_cur, arr_cur;
        new_arr_cur = arr_cur = 0;
        
        while (arr_cur < n) {
            A[new_arr_cur++] = A[arr_cur]; // assign
            //skip duplicates
            while (arr_cur < n && A[arr_cur] == A[new_arr_cur - 1]) {
                arr_cur++;
            }
        }
        
        return new_arr_cur;
    }
};

这里是最多允许两次重复，要在上面的代码进行修改。

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        int new_arr_cur, arr_cur, count;
        new_arr_cur = arr_cur = 0;
        const int DUPLICATE_NUM = 2;
        while (arr_cur < n) {
            A[new_arr_cur++] = A[arr_cur++];
            count = 1;
            //add allowed duplicates
            while (count < DUPLICATE_NUM && arr_cur < n && A[arr_cur] == A[new_arr_cur - 1]) {
                A[new_arr_cur++] = A[arr_cur++];
                count++;
            }
            //skip leftmost duplicate
            while (arr_cur < n && A[arr_cur] == A[new_arr_cur - 1]) {
                arr_cur++;
            }
        }
        return new_arr_cur;
    }
};