【leetcode】1124. Longest Well-Performing Interval

题目如下：

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

解题思路：假设i~j (j不是hours中最后一个元素) 区间是最长符合条件的时间段，那么必定在这个区间内大于8的元素的总数减去小于8的元素的总数的值为1。这里可以通过反证法，如果差值大于1，那么无论第j+1的元素大于还是小于8，i~j+1区间内大于8的元素总数必定是大于小于8的元素的总数的。知道了这个规律就很好办了，遍历hours数组，记录从0开始的0~i区间内大于8的元素的总数与小于8的元素的总数的差值，并记为val[i]，如果j元素是最长区间内最后一个元素，那么只要找到第一个i满足 val[j] - val[i] = 1即可，再判断一下val[j]是否大于0，如果大于0则表示考虑到最长区间是0~j。最后一步再针对最后一个元素单独计算依次即可。

代码如下：

class Solution(object):
    def longestWPI(self, hours):
        """
        :type hours: List[int]
        :rtype: int
        """
        dic = {}
        count = 0
        res = 0
        for i in range(len(hours)-1):
            count = count + 1 if hours[i] > 8 else count - 1
            if count not in dic:
                dic[count] = i
            if count - 1 in dic:
                res = max(res,i - dic[count - 1])
            if count == 1:
                res = max(res,i + 1)

        # the last
        count = 0
        for i in range(len(hours) - 1,-1,-1):
            count = count + 1 if hours[i] > 8 else count - 1
            if count > 0: res = max(res, len(hours) - i)


        return res

标签：count,Performing,res,Interval,1124,dic,hours,大于,元素
来源： https://www.cnblogs.com/seyjs/p/11385853.html