[LeetCode] 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.

基于时间的键值存储。创建一个基于时间的键值存储类 TimeMap，它支持下面两个操作：

1. set(string key, string value, int timestamp)

存储键 key、值 value，以及给定的时间戳 timestamp。
2. get(string key, int timestamp)

返回先前调用 set(key, value, timestamp_prev) 所存储的值，其中 timestamp_prev <= timestamp。
如果有多个这样的值，则返回对应最大的  timestamp_prev 的那个值。
如果没有值，则返回空字符串（""）。

这个题有两种思路，一是用Java的treemap，另一种是用到binary search。

首先第一种思路，用treemap。因为会存在多个键值是同样的key，但是有不同的value和不同的timestamp，所以一开始比较直观的思路一定是hashmap，因为hashmap才能存储类似这种<key, value>键值对的数据。其次，题目要求当找到同样的key，同样的value的时候，要返回最接近timestamp的value，所以想到用treemap，因为treemap是可以自动根据key排序的。

时间O(logn) - treemap的get和set都是这个复杂度

空间O(n)

Java实现

 1 class TimeMap {
 2     private HashMap<String, TreeMap<Integer, String>> map;
 3 
 4     /** Initialize your data structure here. */
 5     public TimeMap() {
 6         map = new HashMap<>();
 7     }
 8 
 9     public void set(String key, String value, int timestamp) {
10         TreeMap<Integer, String> tmap = map.get(key);
11         if (tmap == null) {
12             tmap = new TreeMap<>();
13             map.put(key, tmap);
14         }
15         tmap.put(timestamp, value);
16     }
17 
18     public String get(String key, int timestamp) {
19         TreeMap<Integer, String> tmap = map.get(key);
20         if (tmap == null) {
21             return "";
22         }
23         Integer floorKey = tmap.floorKey(timestamp);
24         if (floorKey == null) {
25             return "";
26         }
27         return tmap.get(floorKey);
28     }
29 }
30 
31 /**
32  * Your TimeMap object will be instantiated and called as such:
33  * TimeMap obj = new TimeMap();
34  * obj.set(key,value,timestamp);
35  * String param_2 = obj.get(key,timestamp);
36  */

第二种思路是用到二分法。在不用treemap的前提下，我们依然需要创建一个hashmap存储key和value（这里会是一个list）的键值对，但是对于value相同但是timestamp不同的数据，我们可以自己创建一个类，把这个类放入list，这样当get key的时候，实际上我们get到的是一个由arraylist串联好的，有相同key的values。当你能得到这个list之后，可以根据给出的变量timestamp找到list中最接近于这个timestamp的键值对。

时间, get - O(1), set - O(logn)

空间O(n)

Java实现

 1 class Data {
 2     String val;
 3     int time;
 4 
 5     Data(String val, int time) {
 6         this.val = val;
 7         this.time = time;
 8     }
 9 }
10 
11 class TimeMap {
12     Map<String, List<Data>> map;
13 
14     /** Initialize your data structure here. */
15     public TimeMap() {
16         map = new HashMap<String, List<Data>>();
17     }
18 
19     public void set(String key, String value, int timestamp) {
20         if (!map.containsKey(key)) {
21             map.put(key, new ArrayList<Data>());
22         }
23         map.get(key).add(new Data(value, timestamp));
24     }
25 
26     public String get(String key, int timestamp) {
27         if (!map.containsKey(key)) {
28             return "";
29         }
30         return binarySearch(map.get(key), timestamp);
31     }
32 
33     private String binarySearch(List<Data> list, int timestamp) {
34         int low = 0;
35         int high = list.size() - 1;
36         while (low < high) {
37             int mid = (low + high) / 2;
38             if (list.get(mid).time == timestamp) {
39                 return list.get(mid).val;
40             }
41             if (list.get(mid).time < timestamp) {
42                 if (list.get(mid + 1).time > timestamp) {
43                     return list.get(mid).val;
44                 }
45                 low = mid + 1;
46             } else {
47                 high = mid - 1;
48             }
49         }
50         return list.get(low).time <= timestamp ? list.get(low).val : "";
51     }
52 }
53 
54 /**
55  * Your TimeMap object will be instantiated and called as such:
56  * TimeMap obj = new TimeMap();
57  * obj.set(key,value,timestamp);
58  * String param_2 = obj.get(key,timestamp);
59  */

LeetCode 题目总结