【leetcode】981. Time Based Key-Value Store

题目如下：

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.

解题思路：我用了两个字典，一个是dic_timestamp，以timestamp为key，value作为val；第二个是dic[val] = [timestamp]，对于val相同的timestamp以升序排列的方式保存在val对应的list中。由于set操作的timestamp是递增的，所以在set的时候只需要把timestamp插入到div[val]的最后即可；对于get操作，采用二分查找的方法找出最大的小于timestamp的历史timestamp。但是二分查找的方法会timeout，后来我发现get操作的timestamp也是递增的，但是题目没有说明，所以改进代码在get操作之后删除掉小于timestamp的所有历史数据。

代码如下：

class TimeMap(object):

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.dic_timestamp = {}
        self.dic = {}


    def set(self, key, value, timestamp):
        """
        :type key: str
        :type value: str
        :type timestamp: int
        :rtype: None
        """
        self.dic[key] = self.dic.setdefault(key,[]) + [timestamp]
        self.dic_timestamp[timestamp] = value

    def get(self, key, timestamp):
        """
        :type key: str
        :type timestamp: int
        :rtype: str
        """
        import bisect
        if key not in self.dic or len(self.dic[key]) == 0 or timestamp < self.dic[key][0]:
            return ''
        elif timestamp >= self.dic[key][-1]:
            v = self.dic[key][-1]
            self.dic[key] = [self.dic[key][-1]]
            return self.dic_timestamp[v]
        v = bisect.bisect_left(self.dic[key], timestamp)
        inx = bisect.bisect_left(self.dic[key], timestamp)
        if inx == len(self.dic[key]) or timestamp != self.dic[key][inx]:
            inx -= 1
        v = self.dic[key][inx]
        self.dic[key] = self.dic[key][inx:]
        return self.dic_timestamp[v]

 

转载于:https://www.cnblogs.com/seyjs/p/10687900.html