LC1052.爱生气的书店老板

思路：滑动窗口

• 分成2部分，第一，记录原有的客户总量all_customers；第二，找到最大的增加量increase_customers，然后计算二者总和即可。
• 问题关键在于，如何寻找最大的增量：可以使用滑动窗口的思想
• 滑动窗口还是分成两部分，第一，找到第一个窗口并计算。第二，将这个窗口向后滑动，每一次扔掉一个左边的数据，加上一个右边的数据，不断更新，直到找到max_increase_customers

代码

class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int minutes) {
        int all_customers = 0, num_customers = customers.size();

        // 记录原有的客户数量
        for (int i = 0; i < num_customers; ++i) {
            all_customers += ((!grumpy[i]) * customers[i]);
        }

        // 统计增加量
        int increase_customers = 0;
        for (int i = 0; i <= minutes - 1; ++i) {
            increase_customers += grumpy[i] * customers[i];
        }

        // 滑动窗口
        int max_increase_customers = increase_customers;
        for (int left = 1, right = minutes; right < num_customers; ++right, ++left) {
            increase_customers -= (grumpy[left - 1] * customers[left - 1]);
            increase_customers += (grumpy[right] * customers[right]);
            max_increase_customers = max(max_increase_customers, increase_customers);
        }

        return all_customers + max_increase_customers;
    }
};