[LeetCode] 1052. Grumpy Bookstore Owner

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

爱生气的书店老板。

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/grumpy-bookstore-owner
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是滑动窗口。这个题我们可以这样想，假设老板一直都不生气，那么我们可以使得所有的customer都满意，那么结果自然是customer数组的累加和。但是现在当老板在某些时候会生气的情况下，我们需要找一个长度为X的子数组，能涵盖尽量多的 grumpy[i] = 1 （老板生气的情况）。所以一开始滑动窗口的end指针往前走，如果end指针指向的index那一刻老板是生气的，则记录一个curSum表示目前能挽回的customer数量；当end - start + 1 == X的时候，说明需要缩短窗口了，此时如果start指针指向的下标老板是生气的，我们则需要把这些customer还回去，因为他们不在窗口里了，也就无法挽回了。

时间O(n)

空间O(1)

Java实现

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int X) {
        int satisfied = 0;
        int len = grumpy.length;
        // 不生气的时候的顾客总数
        for (int i = 0; i < len; i++) {
            if (grumpy[i] == 0) {
                satisfied += customers[i];
            }
        }

        int start = 0;
        int end = 0;
        int curSum = 0;
        int max = 0;
        while (end < len) {
            if (grumpy[end] == 1) {
                curSum += customers[end];
            }
            // if (end - start + 1 >= X)
            if (end - start + 1 == X) {
                max = Math.max(max, curSum);
                if (grumpy[start] == 1) {
                    curSum -= customers[start];
                }
                start++;
            }
            end++;
        }
        return satisfied + max;
    }
}

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