1052.爱生气的书店老板

链接：1052.爱生气的书店老板
难度：Medium
标签：数组、滑动窗口
简介：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。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。

题解 1 - typescript

• 编辑时间：2021-02-23
• 执行用时：104ms
• 内存消耗：44.3MB
• 编程语言：typescript
• 解法介绍：读取范围内生气的值的最大值。

function maxSatisfied(customers: number[], grumpy: number[], X: number): number {
  const _sum = (arr: number[]) => arr.reduce((total, cur) => total + cur, 0);
  if (X === 0) return _sum(customers.filter((_, i) => grumpy[i]));
  if (grumpy.every(v => !v)) return _sum(customers);
  const len = customers.length;
  let l = 0;
  let r = X - 1;
  let sum = _sum(customers.slice(0, X).filter((_, i) => grumpy[i]));
  let maxSum = sum;
  while (r < len - 1) {
    if (grumpy[++r]) sum += customers[r];
    if (grumpy[l]) sum -= customers[l];
    l++;
    maxSum = Math.max(maxSum, sum);
  }
  return maxSum + _sum(customers.filter((_, i) => !grumpy[i]));
}

题解 2 - python

• 编辑时间：2024-04-23
• 执行用时：58ms
• 内存消耗：18.13MB
• 编程语言：python
• 解法介绍：滑动窗口。

class Solution:
    def maxSatisfied(self, customers: List[int], grumpy: List[int], minutes: int) -> int:
        n = len(customers)
        res = cur = sum(customers[i] for i in range(n) if not grumpy[i] or i < minutes)
        for i in range(minutes, n):
            if grumpy[i - minutes]: cur -= customers[i - minutes]
            if grumpy[i]: cur += customers[i]
            res = max(res, cur)
        return res