[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<String, timestamp>，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<timestamp, String>，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

class TimeMap {
    private HashMap<String, TreeMap<Integer, String>> map;

    /** Initialize your data structure here. */
    public TimeMap() {
        map = new HashMap<>();
    }

    public void set(String key, String value, int timestamp) {
        TreeMap<Integer, String> tmap = map.get(key);
        if (tmap == null) {
            tmap = new TreeMap<>();
            map.put(key, tmap);
        }
        tmap.put(timestamp, value);
    }

    public String get(String key, int timestamp) {
        TreeMap<Integer, String> tmap = map.get(key);
        if (tmap == null) {
            return "";
        }
        Integer floorKey = tmap.floorKey(timestamp);
        if (floorKey == null) {
            return "";
        }
        return tmap.get(floorKey);
    }
}

/**
 * Your TimeMap object will be instantiated and called as such:
 * TimeMap obj = new TimeMap();
 * obj.set(key,value,timestamp);
 * String param_2 = obj.get(key,timestamp);
 */

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

class Data {
    String val;
    int time;

    Data(String val, int time) {
        this.val = val;
        this.time = time;
    }
}

class TimeMap {
    Map<String, List<Data>> map;

    /** Initialize your data structure here. */
    public TimeMap() {
        map = new HashMap<String, List<Data>>();
    }

    public void set(String key, String value, int timestamp) {
        if (!map.containsKey(key)) {
            map.put(key, new ArrayList<Data>());
        }
        map.get(key).add(new Data(value, timestamp));
    }

    public String get(String key, int timestamp) {
        if (!map.containsKey(key)) {
            return "";
        }
        return binarySearch(map.get(key), timestamp);
    }

    private String binarySearch(List<Data> list, int timestamp) {
        int low = 0;
        int high = list.size() - 1;
        while (low < high) {
            int mid = (low + high) / 2;
            if (list.get(mid).time == timestamp) {
                return list.get(mid).val;
            }
            if (list.get(mid).time < timestamp) {
                if (list.get(mid + 1).time > timestamp) {
                    return list.get(mid).val;
                }
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return list.get(low).time <= timestamp ? list.get(low).val : "";
    }
}

/**
 * Your TimeMap object will be instantiated and called as such:
 * TimeMap obj = new TimeMap();
 * obj.set(key,value,timestamp);
 * String param_2 = obj.get(key,timestamp);
 */

LeetCode 题目总结