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  • 【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]
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  • 原文地址:https://www.cnblogs.com/seyjs/p/10687900.html
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