[LeetCode] 981. Time Based Key-Value Store

Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key's value at a certain timestamp.

Implement the TimeMap class:

• TimeMap() Initializes the object of the data structure.
• void set(String key, String value, int timestamp) Stores the key key with the value value at the given time timestamp.
• String get(String key, int timestamp) Returns a value such that set was called previously, with timestamp_prev <= timestamp. If there are multiple such values, it returns the value associated with the largest timestamp_prev. If there are no values, it returns "".

Example 1:

Input
["TimeMap", "set", "get", "get", "set", "get", "get"]
[[], ["foo", "bar", 1], ["foo", 1], ["foo", 3], ["foo", "bar2", 4], ["foo", 4], ["foo", 5]]
Output
[null, null, "bar", "bar", null, "bar2", "bar2"]

Explanation
TimeMap timeMap = new TimeMap();
timeMap.set("foo", "bar", 1);  // store the key "foo" and value "bar" along with timestamp = 1.
timeMap.get("foo", 1);         // return "bar"
timeMap.get("foo", 3);         // return "bar", since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 is "bar".
timeMap.set("foo", "bar2", 4); // store the key "foo" and value "bar2" along with timestamp = 4.
timeMap.get("foo", 4);         // return "bar2"
timeMap.get("foo", 5);         // return "bar2"

Constraints:

• 1 <= key.length, value.length <= 100
• key and value consist of lowercase English letters and digits.
• 1 <= timestamp <= 107
• All the timestamps timestamp of set are strictly increasing.
• At most 2 * 105 calls will be made to set and get.

基于时间的键值存储。

设计一个基于时间的键值数据结构，该结构可以在不同时间戳存储对应同一个键的多个值，并针对特定时间戳检索键对应的值。

实现 TimeMap 类：

TimeMap() 初始化数据结构对象
void set(String key, String value, int timestamp) 存储键 key、值 value，以及给定的时间戳 timestamp。
String 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（""）。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/time-based-key-value-store
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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);
 */

 

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