【leetcode】1052. Grumpy Bookstore Owner

题目如下：

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

解题思路：对于grumpy[i] = 0的时间点，显然客户都是满意的，遍历customers可以把所有这些客户的数量求和，同时把对于的customers[i] 设成0。例如示例1中的customers = [1,0,1,2,1,1,7,5]，经过操作后变成 [0,0,0,2,0,1,0,5]。接下来就是问题就转换成求customers中长度为X的连续子数组的和的最大值，最后把第一步的中的和加上第二部操作中的最大值即为结果。

代码如下：

class Solution(object):
    def maxSatisfied(self, customers, grumpy, X):
        """
        :type customers: List[int]
        :type grumpy: List[int]
        :type X: int
        :rtype: int
        """
        res = 0
        for i in range(len(customers)):
            if grumpy[i] == 0:
                res += customers[i]
                customers[i] = 0
        max_count = 0
        val = 0
        for i in range(len(customers)):
            if i < X:
                val += customers[i]
                continue
            max_count = max(val,max_count)
            val -= customers[i-X]
            val += customers[i]
        max_count = max(val, max_count)
        #print customers
        #print res
        return res + max_count