# [[Leetcode]981. Time Based Key-Value Store](https://leetcode.com/problems/time-based-key-value-store/)
- 本题难度: Medium
# Description
Create a timebased key-value store class TimeMap
, that supports two operations.
1. set(string key, string value, int timestamp)
key
and value
, along with the given timestamp
.2. get(string key, int timestamp)
set(key, value, timestamp_prev)
was called previously, with timestamp_prev <= timestamp
.timestamp_prev
.""
).
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, 100]
timestamps
for all TimeMap.set
operations are strictly increasing.1 <= timestamp <= 10^7
TimeMap.set
and TimeMap.get
functions will be called a total of 120000
times (combined) per test case.# 我的代码
```python
import heapq
class TimeMap:
def __init__(self):
"""
Initialize your data structure here.
"""
self.Stores = {}
def set(self, key: str, value: str, timestamp: int) -> None:
tmp = self.Stores.get(key, [])
heapq.heappush(tmp,[timestamp, value])
self.Stores[key] = tmp
def get(self, key: str, timestamp: int) -> str:
nums = self.Stores.get(key, [])
if len(nums) == 0 or nums[0][0]>timestamp:
return ""
left = 0
right = len(nums) - 1
mid = (left + right) // 2
while (left < right):
if nums[mid][0] == timestamp or (
mid < len(nums) - 1 and nums[mid][0] < timestamp and nums[mid + 1][0] > timestamp):
return nums[mid][1]
if nums[mid][0] < timestamp:
left = mid + 1
else:
right = mid - 1
mid = (left + right) // 2
if mid == len(nums) - 1 and nums[mid][0] <= timestamp:
return nums[mid][1]
return ""
```