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【leetcode】1348. Tweet Counts Per Frequency

时间:2020-02-09 18:10:45      阅读:50      评论:0      收藏:0      [点我收藏+]

标签:input   block   represent   exp   time   instant   const   str   turn   

题目如下:

Implement the class TweetCounts that supports two methods:

1. recordTweet(string tweetName, int time)

  • Stores the tweetName at the recorded time (in seconds).

2. getTweetCountsPerFrequency(string freq, string tweetName, int startTime, int endTime)

  • Returns the total number of occurrences for the given tweetName per minute, hour, or day (depending on freq) starting from the startTime (in seconds) and ending at the endTime (in seconds).
  • freq is always minutehour or day, representing the time interval to get the total number of occurrences for the given tweetName.
  • The first time interval always starts from the startTime, so the time intervals are [startTime, startTime + delta*1>,  [startTime + delta*1, startTime + delta*2>, [startTime + delta*2, startTime + delta*3>, ... , [startTime + delta*i, min(startTime + delta*(i+1), endTime + 1)> for some non-negative number i and delta (which depends on freq).  

Example:

Input
["TweetCounts","recordTweet","recordTweet","recordTweet","getTweetCountsPerFrequency","getTweetCountsPerFrequency","recordTweet","getTweetCountsPerFrequency"]
[[],["tweet3",0],["tweet3",60],["tweet3",10],["minute","tweet3",0,59],["minute","tweet3",0,60],["tweet3",120],["hour","tweet3",0,210]]

Output
[null,null,null,null,[2],[2,1],null,[4]]

Explanation
TweetCounts tweetCounts = new TweetCounts();
tweetCounts.recordTweet("tweet3", 0);
tweetCounts.recordTweet("tweet3", 60);
tweetCounts.recordTweet("tweet3", 10);                             // All tweets correspond to "tweet3" with recorded times at 0, 10 and 60.
tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 59); // return [2]. The frequency is per minute (60 seconds), so there is one interval of time: 1) [0, 60> - > 2 tweets.
tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 60); // return [2, 1]. The frequency is per minute (60 seconds), so there are two intervals of time: 1) [0, 60> - > 2 tweets, and 2) [60,61> - > 1 tweet.
tweetCounts.recordTweet("tweet3", 120);                            // All tweets correspond to "tweet3" with recorded times at 0, 10, 60 and 120.
tweetCounts.getTweetCountsPerFrequency("hour", "tweet3", 0, 210);  // return [4]. The frequency is per hour (3600 seconds), so there is one interval of time: 1) [0, 211> - > 4 tweets.

Constraints:

  • There will be at most 10000 operations considering both recordTweet and getTweetCountsPerFrequency.
  • 0 <= time, startTime, endTime <= 10^9
  • 0 <= endTime - startTime <= 10^4

解题思路:题目不难,用hash表就行了,以tweetName作为key值,把所有的time按升级排列保存到key值所对应的list中。查找的时候使用二分查找,找出左右两边的inx即可。

代码如下:

import bisect
class TweetCounts(object):
    dic = {}
    def __init__(self):
        self.dic = {}
        

    def recordTweet(self, tweetName, time):
        """
        :type tweetName: str
        :type time: int
        :rtype: None
        """
        if tweetName not in self.dic:
            self.dic[tweetName] = [time]
        else:
            bisect.insort_left(self.dic[tweetName] ,time)
        

    def getTweetCountsPerFrequency(self, freq, tweetName, startTime, endTime):
        """
        :type freq: str
        :type tweetName: str
        :type startTime: int
        :type endTime: int
        :rtype: List[int]
        """
        if tweetName not in self.dic:
            return []
        res = []
        val = 0
        if freq == minute:
            val = 60
        elif freq == hour:
            val = 3600
        else:val = 86400
        
        start = startTime
        while start <= endTime:
            inx1 = bisect.bisect_left(self.dic[tweetName],start)
            end = min(start + val- 1,endTime)
            inx2 = bisect.bisect_left(self.dic[tweetName],end)
            v = inx2 - inx1
            if inx2 < len(self.dic[tweetName]) and (self.dic[tweetName][inx2] == end) :
                v += 1
            res.append(v)
            start += val
        
        return res


# Your TweetCounts object will be instantiated and called as such:
# obj = TweetCounts()
# obj.recordTweet(tweetName,time)
# param_2 = obj.getTweetCountsPerFrequency(freq,tweetName,startTime,endTime)

 

【leetcode】1348. Tweet Counts Per Frequency

标签:input   block   represent   exp   time   instant   const   str   turn   

原文地址:https://www.cnblogs.com/seyjs/p/12287763.html

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