题目：

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：

与http://blog.csdn.net/lanxu_yy/article/details/11694125类似。方法1：遍历的时候记录重复的元素次数，如果重复则跳过。记录重复元素的方法可以用包含数值的Hash表或者两个BitMap。由于要保持输出数组的有序性，当非重复数值的时候我们尽量将数值前移。方法2：快慢索引。

代码：

方法1：

class bitmap {
    public: bitmap(int n);~bitmap();
    void set(int k);
    bool val(int k);


    private: int len;
    unsigned int * map;
    unsigned int * reversemap;
    bool zero;
};


bitmap::bitmap(int n) {
    len = n / 32 + 1;
    map = new unsigned int[len];
    reversemap = new unsigned int[len];


    for (int i = 0; i < len; i++) {
        map[i] = 0;
        reversemap[i] = 0;
        zero = false;
    }
}


void bitmap::set(int k) {
    bool reverse = true;
    if (k == 0) {
        zero = true;
    } else if (k < 0) {
        reverse = false;
        k = -k;
    }
    int a = k / 32;
    int b = k % 32;


    if (a >= len) {
        return;
    } else {
        if (reverse) {
            int tmp = reversemap[a] / pow(2, b);
            if (tmp % 2 == 0) {
                reversemap[a] += pow(2, b);
            }
        } else {
            int tmp = map[a] / pow(2, b);
            if (tmp % 2 == 0) {
                map[a] += pow(2, b);
            }
        }
    }
}


bool bitmap::val(int k) {
    bool reverse = true;
    if (k == 0) {
        return zero;
    } else if (k < 0) {
        reverse = false;
        k = -k;
    }
    int a = k / 32;
    int b = k % 32;


    if (a >= len) {
        return false;
    } else {
        if (reverse) {
            int tmp = reversemap[a] / pow(2, b);
            if (tmp % 2 == 1) {
                return true;
            } else {
                return false;
            }
        } else {
            int tmp = map[a] / pow(2, b);
            if (tmp % 2 == 1) {
                return true;
            } else {
                return false;
            }
        }
    }
}
class Solution {
public:
    int removeDuplicates(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        bitmap* exist = new bitmap(32767);
        bitmap* duplicate = new bitmap(32767);
        
        int count=0;
        int skip = 0;
        for(int i = 0; i < n; i++)
        {
            if(duplicate->val(A[i]))
            {
                skip++;
                count++;
            }
            else if(exist->val(A[i]))
            {
                duplicate->set(A[i]);
                if(skip > 0)
                {
                    A[i-skip] = A[i];
                }
            }
            else
            {
                exist->set(A[i]);
                if(skip > 0)
                {
                    A[i-skip] = A[i];
                }
            }
            
        }
        return n-count;
    }
};

方法2：

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if(n == 0){
            return 0;
        }
        
        int slow = 0;
        int fast = 1;
        
        while(fast < n){
            if(slow > 1 && A[slow-2] != A[fast] && A[slow-1] == A[fast] && A[slow] == A[fast])
            {
                // ignore
            }
            else if(slow == 1 && A[slow-1] == A[fast] && A[slow] == A[fast]){
                // ignore
            }
            else{
                A[++slow] = A[fast];
            }
            fast++;
        }
        
        return slow + 1;
    }
};