981. Time Based Key-Value Store 基于时间的键值存储


作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/


@TOC

题目地址:https://leetcode.com/problems/time-based-key-value-store/

题目描述

Create a timebased key-value store class TimeMap, that supports two operations.

  1. set(string key, string value, int timestamp)
  • Stores the key and value, along with the given timestamp.
  1. get(string key, int timestamp)
  • Returns a value such that set(key, value, timestamp_prev) was called previously, with timestamp_prev <= timestamp.
  • If there are multiple such values, it returns the one with the largest timestamp_prev.
  • If there are no values, it returns the empty string ("").

Example 1:

Input: inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]
Output: [null,null,"bar","bar",null,"bar2","bar2"]
Explanation:   
TimeMap kv;   
kv.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1   
kv.get("foo", 1);  // output "bar"   
kv.get("foo", 3); // output "bar" since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 ie "bar"   
kv.set("foo", "bar2", 4);   
kv.get("foo", 4); // output "bar2"   
kv.get("foo", 5); //output "bar2"   

Example 2:

Input: inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
Output: [null,null,null,"","high","high","low","low"]

Note:

  1. All key/value strings are lowercase.
  2. All key/value strings have length in the range [1, 100]
  3. The timestamps for all TimeMap.set operations are strictly increasing.
  4. 1 <= timestamp <= 10^7
  5. TimeMap.set and TimeMap.get functions will be called a total of 120000 times (combined) per test case.

题目大意

构建一个带有时间版本的KV存储器。即每次保存的时候会保存当前的时间,查询的时候给出一个时间,要求找到先于该时间的最新的key对应的value。

解题方法

字典

没想到LC还会出这么新颖的题目,这个应用场景在数据库设计中确实会用到。

首先分析一下,我们应该怎么办。首先,如果给出了任意的时间,我们都要找出先于这个时间的key对应的value,说明我们必须把每次的插入结果与对应的时间都保存,而不能使用覆盖的方式。

很显然我们会使用字典这种数据结构来存储kv对,为了保存每个key插入的时间,而且要保证最快的查询时间,我分开保存每次插入的key的time和value。为什么这么做有效呢?因为如果我如果使用key : [(time1, value1), (time2, value2)...]这种存储方式,对快速查找是不利的。而使用key : [time1, time2...]key : [value1, value2...]这种存储方式能保证time和value是一一对应的。所以这种方式先根据key和time快速查找到小于该时间的timex,然后就能根据索引快速找到此索引对应的valuex.

在有序列表中快速查找小于一个time的time_x,当然使用二分了,所以使用了bisect_right来快速查找到了一个不大于该time的时间对应的索引,然后拿这个索引找到对应的value即可。

万万没想到的是,竟然没通过!我已经优化了存储和查找啊!看了下没通过的测试用例,发现是一个键值对反复的插入查找,每次查找的时间都是最新的时间。好吧,那么我使用了一个这个max_字典,用来保存当前的key更新的时间。这样的好处是,当我们查找一个不小于当前时间的值的时候,一定是最后一次插入的那个时间。

最后,说句题外话,这个题没有考到的是我们的删除操作怎么办?其实业界做法是使用延迟删除的方式,即插入一个标志位代表删除,而不进行真正的删除操作。比如我们在最后的时刻设置key为"",这样我们查找的时候发现这个key的数值是""就意味着被删除。

class TimeMap(object):

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.t_ = collections.defaultdict(list)
        self.v_ = collections.defaultdict(list)
        self.max_ = collections.defaultdict(int)
        

    def set(self, key, value, timestamp):
        """
        :type key: str
        :type value: str
        :type timestamp: int
        :rtype: None
        """
        self.t_[key].append(timestamp)
        self.v_[key].append(value)
        self.max_[key] = max(self.max_[key], timestamp)
        

    def get(self, key, timestamp):
        """
        :type key: str
        :type timestamp: int
        :rtype: str
        """
        if key not in self.t_:
            return ""
        if timestamp >= self.max_[key]:
            return self.v_[key][-1]
        v = bisect.bisect_right(self.t_[key], timestamp)
        if v:
            return self.v_[key][v - 1]
        return ""


# Your TimeMap object will be instantiated and called as such:
# obj = TimeMap()
# obj.set(key,value,timestamp)
# param_2 = obj.get(key,timestamp)

日期

2019 年 1 月 27 日 —— 这个周赛不太爽