滑动窗口法求Sum

1052. Grumpy Bookstore Owner

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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.

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int X) {
        int Sum = 0;
		int Result = 0;
		for(int i = 0;i < customers.length;i ++) {
			if(i < X) {
				Sum += customers[i];
			}
			else {
				Sum += customers[i] * (1 - grumpy[i]);
			}
		}
		Result = Sum;
		for(int j = 0;j < customers.length - X;j ++) {
			if(grumpy[j] == 1) {
				Sum = Sum - customers[j];
			}
			if(grumpy[j + X] == 1) {
				Sum = Sum + customers[j + X];
			}
			Result = Math.max(Result,Sum);
		}
		return Result;
    }
}