Create a timebased key-value store class TimeMap
, that supports two operations.
1. set(string key, string value, int timestamp)
- Stores the
key
andvalue
, along with the giventimestamp
.
2. get(string key, int timestamp)
- Returns a value such that
set(key, value, timestamp_prev)
was called previously, withtimestamp_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"
思路:这题就是考察,bianry search,用HashMap<String, List<Node>> 把value , time存下来之后,因为是递增的,所以可以做binary search;T: O(NlogN) S: O(N) 注意,有可能搜索的范围根本没有 小于 time 的value,此时要返回null;
class TimeMap {
private class Node {
public String value;
public int time;
public Node(String value, int time) {
this.value = value;
this.time = time;
}
}
private HashMap<String, List<Node>> hashmap;
/** Initialize your data structure here. */
public TimeMap() {
hashmap = new HashMap<>();
}
public void set(String key, String value, int timestamp) {
hashmap.putIfAbsent(key, new ArrayList<Node>());
Node node = new Node(value, timestamp);
hashmap.get(key).add(node);
}
public String get(String key, int timestamp) {
if(!hashmap.containsKey(key)) {
return "";
} else {
// contains; Do binary search on list;
List<Node> list = hashmap.get(key);
Node node = binarySearch(list, timestamp);
return node == null ? "" : node.value;
}
}
private Node binarySearch(List<Node> list, int target) {
int start = 0;
int end = list.size() - 1;
// find first element.time <= target;
while(start + 1 < end) {
int mid = start + (end - start) / 2;
if(list.get(mid).time == target) {
return list.get(mid);
}
if(list.get(mid).time < target) {
start = mid;
} else {
// list.get(mid).time > target
end = mid;
}
}
// 注意有可能 start,end都不小于target,也就是说不在搜索范围; return null;
if(list.get(end).time <= target) {
return list.get(end);
}
if(list.get(start).time <= target) {
return list.get(start);
}
return null;
}
}
/**
* 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);
*/