1386. 安排电影院座位

先sort一下，就能将reservedSeats 数组按照行和列排序，设begin = 0， end = 11

则遍历每个reservedSeats时，将有cnt = reservedSeats[i][1]  - begin - 1个空位

如果reservedSeats[i][0] == reservedSeats[i + 1][0]，则说明在同一行，令begin = reservedSeats[i][0] 即可

否则，begin = 0，并计算出剩余位置有多少空位 即 cnt = end - reservedSeats[i][1] - 1

每次计算出空位时需判断是否在不合适的位置，就10个位置，哪个不合适，自己看一下就行，看代码也能看出来

class Solution {
public:

    int maxNumberOfFamilies(int n, vector<vector<int>>& reservedSeats) {
        
        int len = reservedSeats.size();
        sort(reservedSeats.begin(), reservedSeats.end());
        int row = 1;
        int ret = 0, begin = 0, end = 11;
        for(int i = 0; i < len - 1; i++)
        {

            int cnt = reservedSeats[i][1] - begin - 1;
            if(cnt >= 4)
            {
                if(begin == 0 || begin == 4 || begin == 2 || begin == 6) cnt--;
                if(cnt >= 4) ret += cnt / 4;
            }
            if(reservedSeats[i + 1][0] == reservedSeats[i][0]) begin = reservedSeats[i][1];
            else
            {
                row++;
                begin = 0;
                cnt = end - reservedSeats[i][1] - 1;
                if(cnt >= 4)
                {
                    if(reservedSeats[i][1] == 0 || reservedSeats[i][1] == 4 || reservedSeats[i][1] == 2 || reservedSeats[i][1] == 6) cnt--;
                    if(cnt >= 4) ret += cnt / 4;
                }

            }
        }
        int cnt = reservedSeats[len - 1][1] - begin - 1;
        if(cnt >= 4)
        {
            if(begin == 0 || begin == 4 || begin == 2 || begin == 6) cnt--;
            if(cnt >= 4) ret += cnt / 4;
        }
        cnt = end - reservedSeats[len - 1][1] - 1;

        if(cnt >= 4)
        {
            if(reservedSeats[len - 1][1] == 0 || reservedSeats[len - 1][1] == 4 || reservedSeats[len - 1][1] == 2 || reservedSeats[len - 1][1] == 6) cnt--;
            if(cnt >= 4) ret += cnt / 4;
        }
        ret += (n - row) * 2;



        return ret;




    }
};