1 題目

You are given several logs that each log contains a unique id and timestamp. Timestamp is a string that has the following format: Year:Month:Day:Hour:Minute:Second, for example, 2017:01:01:23:59:59. All domains are zero-padded decimal numbers.
Design a log storage system to implement the following functions:
void Put(int id, string timestamp): Given a log’s unique id and timestamp, store the log in your storage system.

int[] Retrieve(String start, String end, String granularity): Return the id of logs whose timestamps are within the range from start to end. Start and end all have the same format as timestamp. However, granularity means the time level for consideration. For example, start = “2017:01:01:23:59:59”, end = “2017:01:02:23:59:59”, granularity = “Day”, it means that we need to find the logs within the range from Jan. 1st 2017 to Jan. 2nd 2017.

Example 1:

put(1, "2017:01:01:23:59:59");
put(2, "2017:01:01:22:59:59");
put(3, "2016:01:01:00:00:00");
retrieve("2016:01:01:01:01:01","2017:01:01:23:00:00","Year"); // return [1,2,3], because you need to return all logs within 2016 and 2017.
retrieve("2016:01:01:01:01:01","2017:01:01:23:00:00","Hour"); // return [1,2], because you need to
 
 return all logs start from 2016:01:01:01 to 2017:01:01:23, where log 3 is left outside the range.

Note:
1. There will be at most 300 operations of Put or Retrieve.
2. Year ranges from [2000,2017]. Hour ranges from [00,23].
3. Output for Retrieve has no order required.

2 設計方案

日誌系統中：

• 時間戳應該儲存為某一起始時間點（例如1970年1月1日0時0分0秒）以來的秒數。這樣字元型的時間戳被轉化為整型，便於儲存和查詢。
• 時間戳連同日誌記錄的id，以及其他資訊作為一個節點儲存在特定資料結構中（連結串列或者搜尋樹）。

3 程式碼

public class LogSystem {
    ArrayList < long[] > list;
    public LogSystem() {
        list = new ArrayList < long[] > ();
    }

    public void put(int id, String timestamp) {
        int[] st = Arrays.stream(timestamp.split(":")).mapToInt(Integer::parseInt).toArray();
        list.add(new long[] {convert(st), id});
    }
    public long convert(int[] st) {
        st[1] = st[1] - (st[1] == 0 ? 0 : 1);
        st[2] = st[2] - (st[2] == 0 ? 0 : 1);
        return (st[0] - 1999L) * (31 * 12) * 24 * 60 * 60 + st[1] * 31 * 24 * 60 * 60 + st[2] * 24 * 60 * 60 + st[3] * 60 * 60 + st[4] * 60 + st[5];
    }
    public List < Integer > retrieve(String s, String e, String gra) {
        ArrayList < Integer > res = new ArrayList();
        long start = granularity(s, gra, false);
        long end = granularity(e, gra, true);
        for (int i = 0; i < list.size(); i++) {
            if (list.get(i)[0] >= start && list.get(i)[0] < end)
                res.add((int) list.get(i)[1]);
        }
        return res;
    }

    public long granularity(String s, String gra, boolean end) {
        HashMap < String, Integer > h = new HashMap();
        h.put("Year", 0);
        h.put("Month", 1);
        h.put("Day", 2);
        h.put("Hour", 3);
        h.put("Minute", 4);
        h.put("Second", 5);
        String[] res = new String[] {"1999", "00", "00", "00", "00", "00"};
        String[] st = s.split(":");
        for (int i = 0; i <= h.get(gra); i++) {
            res[i] = st[i];
        }
        int[] t = Arrays.stream(res).mapToInt(Integer::parseInt).toArray();
        if (end)
            t[h.get(gra)]++;
        return convert(t);
    }
}

參考文獻