1124. Longest Well-Performing Interval

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分类专栏： leetcode hashtable

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博客围绕给定员工每日工作时长列表，定义了疲劳日和表现良好时间段。目标是求最长表现良好时间段的长度。思路是将工作时长大于8的转化为1，小于8的转化为 -1，把问题转化为求最长子数组使其和大于0，还介绍了利用哈希表处理的方法。

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We are given hours, a list of the number of hours worked per day for a given employee.

A day is considered to be a tiring day if and only if the number of hours worked is (strictly) greater than 8.

A well-performing interval is an interval of days for which the number of tiring days is strictly larger than the number of non-tiring days.

Return the length of the longest well-performing interval.

Example 1:

Input: hours = [9,9,6,0,6,6,9]
Output: 3
Explanation: The longest well-performing interval is [9,9,6].

Constraints:

1 <= hours.length <= 10000
0 <= hours[i] <= 16

思路：>8的转化为1，<8的转化为-1

于是题目就变成求最长的subarray，使得该subarray的和大于0

如果是等于0，直接hashtable查找就行

大于0需要变一下，还是做从0开始的running sum，用hashtable记录负的sum，

假设遍历到index位置，从0开始的running sum为-1，那就从hashtable里面找-2（所以存的时候-2一定要存数组索引最小的，转化到代码就是如果-2已经在hashtable就不更新），这样中间的区间sum就为1

为什么不找-3，-4呢？因为他们不会出现在-2的前面（-2过了才能减到-3，而-2是最优的）

class Solution(object):
    def longestWPI(self, hours):
        """
        :type hours: List[int]
        :rtype: int
        """
        helper = [1 if a>8 else -1 for a in hours]
        d = {}
        res = 0
        running = 0
        for i,a in enumerate(helper):
            running += a
            if running>0:
                res=max(res,i+1)
            else:
                if running-1 in d:
                    res=max(res,i-d[running-1])
                if running not in d: d[running]=i
                
        return res