README.md

1052. 爱生气的书店老板

English Version

题目描述

有一个书店老板，他的书店开了 n 分钟。每分钟都有一些顾客进入这家商店。给定一个长度为 n 的整数数组 customers ，其中 customers[i] 是在第 i 分钟开始时进入商店的顾客数量，所有这些顾客在第 i 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，若老板不生气则顾客是满意的。

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 minutes 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 。
 

示例 1：

输入：customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], minutes = 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.

示例 2：

输入：customers = [1], grumpy = [0], minutes = 1
输出：1

 

提示：

• n == customers.length == grumpy.length
• 1 <= minutes <= n <= 2 * 104
• 0 <= customers[i] <= 1000
• grumpy[i] == 0 or 1

解法

方法一：滑动窗口

定义 $s$ 表示所有不满意的顾客总数，$cs$ 表示顾客总数。

Python3

class Solution:
    def maxSatisfied(
        self, customers: List[int], grumpy: List[int], minutes: int
    ) -> int:
        s = sum(a * b for a, b in zip(customers, grumpy))
        cs = sum(customers)
        t = ans = 0
        for i, (a, b) in enumerate(zip(customers, grumpy), 1):
            t += a * b
            if (j := i - minutes) >= 0:
                ans = max(ans, cs - (s - t))
                t -= customers[j] * grumpy[j]
        return ans

Java

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        int s = 0, cs = 0;
        int n = customers.length;
        for (int i = 0; i < n; ++i) {
            s += customers[i] * grumpy[i];
            cs += customers[i];
        }
        int t = 0, ans = 0;
        for (int i = 0; i < n; ++i) {
            t += customers[i] * grumpy[i];
            int j = i - minutes + 1;
            if (j >= 0) {
                ans = Math.max(ans, cs - (s - t));
                t -= customers[j] * grumpy[j];
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int minutes) {
        int s = 0, cs = 0;
        int n = customers.size();
        for (int i = 0; i < n; ++i) {
            s += customers[i] * grumpy[i];
            cs += customers[i];
        }
        int t = 0, ans = 0;
        for (int i = 0; i < n; ++i) {
            t += customers[i] * grumpy[i];
            int j = i - minutes + 1;
            if (j >= 0) {
                ans = max(ans, cs - (s - t));
                t -= customers[j] * grumpy[j];
            }
        }
        return ans;
    }
};

Go

func maxSatisfied(customers []int, grumpy []int, minutes int) int {
	s, cs := 0, 0
	for i, c := range customers {
		s += c * grumpy[i]
		cs += c
	}
	t, ans := 0, 0
	for i, c := range customers {
		t += c * grumpy[i]
		j := i - minutes + 1
		if j >= 0 {
			ans = max(ans, cs-(s-t))
			t -= customers[j] * grumpy[j]
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Rust

impl Solution {
    pub fn max_satisfied(customers: Vec<i32>, grumpy: Vec<i32>, minutes: i32) -> i32 {
        let k = minutes as usize;
        let n = customers.len();

        let mut sum = 0;
        for i in 0..k {
            if grumpy[i] == 1 {
                sum += customers[i];
            }
        }
        let mut max = sum;
        for i in k..n {
            if grumpy[i - k] == 1 {
                sum -= customers[i - k];
            }
            if grumpy[i] == 1 {
                sum += customers[i];
            }
            max = max.max(sum);
        }

        sum = 0;
        for i in 0..n {
            if grumpy[i] == 0 {
                sum += customers[i];
            }
        }
        sum + max
    }
}

...