981. Time Based Key-Value Store

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.

2. 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.

这道题让我们实现一种基于时间的键值对儿数据结构,有两种操作 set 和 get,其中 set 就是存入键值对儿,同时需要保存时间戳,get 就是查找值,但此时不仅提供了 key 值,还提供了查询的时间戳,返回值的时间戳不能大于查询的时间戳,假如有多个相同值,返回时间戳最大的那个,若查询不到就返回空。实际上这道题考察的就是较为复杂一些的数据结构,因为要同时保存三个量,而且还要提供快速查询功能,可以使用 Map of Maps 的数据结构,外层可以使用一个 HashMap,因为对于 key 值没有顺序要求,而内层要使用一个 TreeMap,因为时间戳的顺序很重要。在 set 函数中直接将数据插入数据结构中,在 get 中,用一个 upper_bound 来进行快速查找第一个大于目标值的位置,往后退一位,就是不大于目标值的位置。但是在退之前要判断得到的位置是否是起始位置,是的话就没法再往前退一位了,直接返回空串,不是的话可以退一位并返回即可,参见代码如下:

class TimeMap {
public:
    TimeMap() {}
    
    void set(string key, string value, int timestamp) {
        dataMap[key].insert({timestamp, value});
    }
    
    string get(string key, int timestamp) {
        auto it = dataMap[key].upper_bound(timestamp);
        return it == dataMap[key].begin() ? "" : prev(it)->second;
    }

private:
    unordered_map<string, map<int, string>> dataMap;
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/981

参考资料:

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

https://leetcode.com/problems/time-based-key-value-store/discuss/226663/TreeMap-Solution-Java

https://leetcode.com/problems/time-based-key-value-store/discuss/226664/C%2B%2B-3-lines-hash-map-%2B-map

LeetCode All in One 题目讲解汇总(持续更新中…)


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