[LeetCode] 1052. Grumpy Bookstore Owner

Today, the bookstore owner has a store open for customers.length minutes. Every minute, some number of customers (customers[i]) enter the store, and all those customers leave after the end of that minute.

On some minutes, the bookstore owner is grumpy. If the bookstore owner is grumpy on the i-th minute, grumpy[i] = 1, otherwise grumpy[i] = 0. When the bookstore owner is grumpy, the customers of that minute are not satisfied, otherwise they are satisfied.

The bookstore owner knows a secret technique to keep themselves not grumpy for X minutes straight, but can only use it once.

Return the maximum number of customers that can be satisfied throughout the day.

Example 1:
Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 3
Output: 16
Explanation: The bookstore owner keeps themselves not grumpy for the last 3 minutes.
The maximum number of customers that can be satisfied = 1 + 1 + 1 + 1 + 7 + 5 = 16.

Note:
1 <= X <= customers.length == grumpy.length <= 20000
0 <= customers[i] <= 1000
0 <= grumpy[i] <= 1

爱生气的书店老板。

今天，书店老板有一家店打算试营业 customers.length 分钟。每分钟都有一些顾客（customers[i]）会进入书店，所有这些顾客都会在那一分钟结束后离开。

在某些时候，书店老板会生气。 如果书店老板在第 i 分钟生气，那么 grumpy[i] = 1，否则 grumpy[i] = 0。 当书店老板生气时，那一分钟的顾客就会不满意，不生气则他们是满意的。

书店老板知道一个秘密技巧，能抑制自己的情绪，可以让自己连续 X 分钟不生气，但却只能使用一次。

请你返回这一天营业下来，最多有多少客户能够感到满意的数量。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/grumpy-bookstore-owner
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思路

思路是滑动窗口。这道题我提供两个做法，一个是固定长度的滑动窗口，一个是类似76题的那种模板。
可以用固定长度做是因为老板有一次机会可以让自己连续 X 分钟不生气。

复杂度

时间O(n)
空间O(1)

代码一 - 固定长度的滑动窗口

Java实现

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        int n = customers.length;
        int sum = 0;
        int k = minutes;
        // 如果前k分钟不生气的满意值
        for (int i = 0; i < k; i++) {
            sum += customers[i];
        }
		// 之后的分钟按是否生气正常累加客户的满意值
        for (int i = k; i < n; i++) {
            if (grumpy[i] == 0) {
                sum += customers[i];
            }
        }

        int res = sum;
        for (int i = k; i < n; i++) {
            if (grumpy[i] == 1) {
                sum += customers[i];
            }
            if (grumpy[i - k] == 1) {
                sum -= customers[i - k];
            }
            res = Math.max(res, sum);
        }
        return res;
    }
}

代码二

Java实现

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        int satisfied = 0;
        int n = customers.length;
		// 老板不生气的时段内顾客的满意度
        for (int i = 0; i < n; i++) {
            if (grumpy[i] == 0) {
                satisfied += customers[i];
            }
        }

        int start = 0;
        int end = 0;
        int curSum = 0;
        int max = 0;
        while (end < n) {
            if (grumpy[end] == 1) {
                curSum += customers[end];
            }
            end++;
			// 如果长度正好等于minutes，统计当前max的值，代表能有多少顾客的满意度可以被改变
            if (end - start == minutes) {
                max = Math.max(max, curSum);
                if (grumpy[start] == 1) {
                    curSum -= customers[start];
                }
                start++;
            }
        }
        return satisfied + max;
    }
}

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