981. Time Based Key-Value Store #

题目 #

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.

题目大意 #

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（""）。

提示：

1. 所有的键/值字符串都是小写的。
2. 所有的键/值字符串长度都在 [1, 100] 范围内。
3. 所有 TimeMap.set 操作中的时间戳 timestamps 都是严格递增的。
4. 1 <= timestamp <= 10^7
5. TimeMap.set 和 TimeMap.get 函数在每个测试用例中将（组合）调用总计 120000 次。

解题思路 #

• 要求设计一个基于时间戳的 kv 存储。set() 操作里面会会包含一个时间戳。get() 操作的时候查找时间戳小于等于 timestamp 的 key 对应的 value，如果有多个解，输出满足条件的最大时间戳对应的 value 值。
• 这一题可以用二分搜索来解答，用 map 存储 kv 数据，key 对应的就是 key，value 对应一个结构体，里面包含 value 和 timestamp。执行 get() 操作的时候，先取出 key 对应的结构体数组，然后在这个数组里面根据 timestamp 进行二分搜索。由于题意是要找小于等于 timestamp 的最大 timestamp ，这会有很多满足条件的解，变换一下，先找 > timestamp 的最小解，然后下标减一即是满足题意的最大解。
• 另外题目中提到“TimeMap.set 操作中的 timestamp 是严格递增的”。所以在 map 中存储 value 结构体的时候，不需要排序了，天然有序。

代码 #

package leetcode

import "sort"

type data struct {
	time  int
	value string
}

// TimeMap is a timebased key-value store
// TimeMap define
type TimeMap map[string][]data

// Constructor981 define
func Constructor981() TimeMap {
	return make(map[string][]data, 1024)
}

// Set define
func (t TimeMap) Set(key string, value string, timestamp int) {
	if _, ok := t[key]; !ok {
		t[key] = make([]data, 1, 1024)
	}
	t[key] = append(t[key], data{
		time:  timestamp,
		value: value,
	})
}

// Get define
func (t TimeMap) Get(key string, timestamp int) string {
	d := t[key]
	i := sort.Search(len(d), func(i int) bool {
		return timestamp < d[i].time
	})
	i--
	return t[key][i].value
}

/**
 * Your TimeMap object will be instantiated and called as such:
 * obj := Constructor();
 * obj.Set(key,value,timestamp);
 * param_2 := obj.Get(key,timestamp);
 */