Rust 力扣 - 1052. 爱生气的书店老板

题目描述

题解思路

我们遍历长度为minutes的窗口，我们需要记录窗口内的生气覆盖的客户数量，同时记录全局不生气覆盖的客户数量，遍历过程中刷新窗口内的生气覆盖的客户数量最大值

结果为全局不生气覆盖的客户数量 + 窗口内的生气覆盖的客户数量最大值

题解代码

impl Solution {
	pub fn max_satisfied(customers: Vec<i32>, grumpy: Vec<i32>, minutes: i32) -> i32 {
	    let mut sum = 0;
	    let mut ans = sum;
	
	    for i in 0..minutes as usize {
	        if grumpy[i] == 0 {
	            ans += customers[i];
	        } else {
	            sum += customers[i];
	        }
	    }
	
	    let mut max_sum = sum;
	
	    for i in minutes as usize..customers.len() {    
	        if grumpy[i] == 0 {
	            ans += customers[i];
	        }
	
	        sum += customers[i] * grumpy[i] - customers[i - minutes as usize] * grumpy[i - minutes as usize];
	        max_sum = max_sum.max(sum);
	    }
	
	    ans + max_sum
	}
}

题目链接

https://leetcode.cn/problems/grumpy-bookstore-owner/description/