[Swift]LeetCode1052.爱生气的书店老板 | Grumpy Bookstore Owner

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➤微信公众号：山青咏芝（shanqingyongzhi）
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➤原文地址：https://www.cnblogs.com/strengthen/p/10925087.html 
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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

---

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。

示例：

输入：customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 3
输出：16
解释：
书店老板在最后 3 分钟保持冷静。
感到满意的最大客户数量 = 1 + 1 + 1 + 1 + 7 + 5 = 16.

提示：

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

---

240ms

class Solution {
    func maxSatisfied(_ customers: [Int], _ grumpy: [Int], _ X: Int) -> Int {
        var base = 0
        for i in 0..<customers.count where grumpy[i] == 0 {
            base += customers[i]
        }
        
        var curr = base, ans = base
        for i in 0..<customers.count {
            if grumpy[i] == 1 {
                curr += customers[i]
            }

            if i >= X {
                if grumpy[i-X] == 1 {
                    curr -= customers[i-X]
                }
            }
            ans = max(ans, curr)
        }
        
        return ans
    }
}

---

Runtime: 244 ms

Memory Usage: 21.3 MB

class Solution {
    func maxSatisfied(_ customers: [Int], _ grumpy: [Int], _ X: Int) -> Int {
        var customers:[Int] = customers
        var grumpy:[Int] = grumpy
        let n:Int = customers.count
        
        var ans:Int = 0
        for i in 0..<n
        {
            if (grumpy[i] == 0) || i < X
            {
                ans += customers[i]
            }
        }
        var cur:Int = ans
        for p in X..<n
        {
            let q:Int = p - X
            if grumpy[q] != 0 {cur -= customers[q]}
            if grumpy[p] != 0 {cur += customers[p]}
            ans = max(ans, cur)
        }
        return ans
    }
}