635. Design Log Storage System

635. Design Log Storage System

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.

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Solution #1, log都存在list里面，每次都要遍历
O(n), n是总大小

class LogSystem {

    private List<String[]> logs;
    private Map<String, Integer> map;
    
    public LogSystem() {
        logs = new ArrayList<>();
        map = new HashMap<>();
        map.put("Year", 4);
        map.put("Month", 7);
        map.put("Day", 10);
        map.put("Hour", 13);
        map.put("Minute", 16);
        map.put("Second", 19);
    }
    
    public void put(int id, String timestamp) {
        String[] log = new String[]{Integer.toString(id), timestamp};
        logs.add(log);
    }
    
    public List<Integer> retrieve(String s, String e, String gra) {
        int len = map.get(gra);
        List<Integer> rt = new ArrayList<>();
        for (String[] log : logs) {
            if (log[1].substring(0,len).compareTo(s.substring(0,len)) >= 0
               && log[1].substring(0,len).compareTo(e.substring(0,len)) <= 0) {
                rt.add(Integer.parseInt(log[0]));
            }
        }
        
        return rt;
    }
}

/**
 * Your LogSystem object will be instantiated and called as such:
 * LogSystem obj = new LogSystem();
 * obj.put(id,timestamp);
 * List<Integer> param_2 = obj.retrieve(s,e,gra);
 */

Solution #2用treemap来存log，然后范围搜索用subMap，注意OJ会报编译错误，但Intellij不会
O(n), n是范围内的log数量

class LogSystem {

    private TreeMap<String, List<Integer>> logs;
    private Map<String, Integer> map;
    private String min;
    private String max;
    
    public LogSystem() {
        min = "2000:01:01:00:00:00";
        max = "2017:12:31:23:59:59";
        
        logs = new TreeMap<>();
        map = new HashMap<>();
        map.put("Year", 4);
        map.put("Month", 7);
        map.put("Day", 10);
        map.put("Hour", 13);
        map.put("Minute", 16);
        map.put("Second", 19);
    }
    
    public void put(int id, String timestamp) {
        String[] log = new String[]{Integer.toString(id), timestamp};
        List<Integer> l = logs.getOrDefault(timestamp, new ArrayList<Integer>());
        l.add(id);
        logs.put(timestamp, l);
    }
    
    public List<Integer> retrieve(String s, String e, String gra) {
        int len = map.get(gra);
        NavigableMap<String, List<Integer>> sub = logs.subMap(s.substring(0,len) + min.substring(len),
                                                               true,
                                                               e.substring(0,len) + max.substring(len),
                                                               true);
        
        List<Integer> rt = new ArrayList<>();
        for (Map.Entry<String, List<Integer>> en : sub.entrySet()) {
            rt.addAll(en.getValue());
        }
        
        return rt;
    }
}

Solution #3, Trie, 每一层存的是year,month,day...
跟#2的复杂度是一样，但是写起来更麻烦