# 1052.Grumpy-Bookstore-Owner

## 题目地址

https://leetcode.com/problems/grumpy-bookstore-owner/

## 题目描述

``````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``````

## 代码

### Approach #1 Sliding Window

``````class Solution {
public int maxSatisfied(int[] customers, int[] grumpy, int X) {
// 1. 计算原本就开心的客户数量 n base
int baseCount = 0;
for (int i = 0; i < customers.length; i++) {
if (grumpy[i] == 0) {
baseCount += customers[i];
}
}

// 2. 生成 不开心客户的数组
for (int i = 0; i < customers.length; i++) {
if (grumpy[i] == 1) {
}
}

// 3. 计算 X 分钟板蓝根 所带来的最大收益
// 3.1 先计算 前 X 分钟的收益
int gain = 0;
for (int i = 0; i < X; i++) {
}

// 3.2 采用 改变量 计算接下来每个区间的 收益并 记录最大值
int maxGain = gain;
for (int i = X; i < sadCustomers.length; i++) {
if (gain > maxGain) {
maxGain = gain;
}
}

// 4. 求和
return baseCount + maxGain;
}
}``````

### Approach #2

``````class Solution {
public int maxSatisfied(int[] customers, int[] grumpy, int X) {
int maxGrumpy = 0;
int happy = 0;
int g = 0;
for (int i = 0; i < customers.length; i++) {
if (i >= X) {
g = g - (customers[i - X] * grumpy[i - X]);
}
g += (customers[i] * grumpy[i]);
maxGrumpy = Math.max(maxGrumpy, g);
if (grumpy[i] == 0) {
happy += customers[i];
}
}

happy += maxGrumpy;
return happy;
}
}``````

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