Leetcode 1052 Grumpy Bookstore Owner. (滑动窗口)
问题描述
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.
例子
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.
方法
1. 使用滑动窗口windows记录不满意的客户数（X分钟）。当滑动窗的宽度大于X时从滑动窗的左端减去不满意的客户
windows -= grumpy[i - X] * customers[i - X];
2. 使用satisfied记录satistified客户数量没有脾气暴躁的技术;
3. 在迭代结束时，satisfied+ max(winOfMakeSatisfied)是答案。
** Solution Java **
** 5ms, beats 24.57% **
** 52.9MB, beats 100.00% **
class Solution {
public int maxSatisfied(int[] customers, int[] grumpy, int X) {
int satisfied = 0, maxMakeSatisfied = 0, windows = 0;
for (int i = 0; i < customers.length; ++i) {
if (grumpy[i] == 0)
satisfied += customers[i];
else
windows += customers[i];
if (X <= i) {
windows -= grumpy[i - X] * customers[i - X];
}
maxMakeSatisfied = Math.max(maxMakeSatisfied, windows);
}
return satisfied + maxMakeSatisfied;
}
}
** Solution Python3 **
** 320ms, beats 42.52% **
** 15MB, beats 100.00% **
class Solution:
def maxSatisfied(self, customers: List[int], grumpy: List[int], X: int) -> int:
i = win_of_make_satisfied = satisfied = max_make_satisfied = 0
for c, g in zip(customers, grumpy):
satisfied += (1 - g) * c
win_of_make_satisfied += g * c
if i >= X:
win_of_make_satisfied -= grumpy[i - X] * customers[i - X]
max_make_satisfied = max(win_of_make_satisfied, max_make_satisfied)
i += 1
return satisfied + max_make_satisfied
