[Leetcode] 635. Design Log Storage System 解题报告

题目：

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:

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

我第一眼看到题目的时候，觉得应该用线段树，后来发现直接用简单的map就可以使得put函数和retrieve函数的时间复杂度都达到O(logn)。具体做法是：用时间戳（timestamp）作为map的key，用id作为map的value。这样给定一个开始时间和一个结束时间后，我们只要输出两者之间的所有id即可。

算法就这么简单，但实现中需要注意如何处理timestamp。我的做法是用vector<int>来作为key。在retrieve的时候，我们需要根据gra来调整key的值：在gra之后的int值，如果表示开始，则统统置为0；否则就置为该时间粒度上的最大值（例如月份的最大值是12，天数的最大值是31，时间
b396
的最大值是23，分和秒的最大值是59）。这样我们只要用lower_bound和upper_bound就可以bound正确的区间了。

代码：

class LogSystem {
public:
LogSystem() {

}

void put(int id, string timestamp) {
vector<int> time = getTime(timestamp, "Second", true);
mp[time] = id;
}

vector<int> retrieve(string s, string e, string gra) {
vector<int> start_time = getTime(s, gra, false);
vector<int> end_time = getTime(e, gra, true);
map<vector<int>, int>::iterator it1 = mp.lower_bound(start_time);
map<vector<int>, int>::iterator it2 = mp.upper_bound(end_time);
vector<int> ret;
for (; it1 != it2; ++it1) {
ret.push_back(it1->second);
}
return ret;
}
private:
vector<int> getTime(const string &s, const string &gra, bool is_end) const {
int index = getIndex(gra);
istringstream iss(s);
vector<int> time(6, 0);
int value = 0;
char c = ':';
for (int i = 0; i <= index; ++i) {
iss >> value;
time[i] = value;
iss >> c;
}
if (is_end) {
for (int i = index + 1; i <= 5; ++i) {
time[i] = getMax(i);
}
}
return time;
}
int getIndex(const string &gra) const {
if (gra == "Year") {
return 0;
}
else if (gra == "Month") {
return 1;
}
else if (gra == "Day") {
return 2;
}
else if (gra == "Hour") {
return 3;
}
else if (gra == "Minute") {
return 4;
}
else {
return 5;
}
}
int getMax(int index) const {
switch(index) {
case 1: // month
return 12;
case 2: // day (not necessary the accurate)
return 31;
case 3: // hour
return 23;
case 4: // minute
return 59;
default: // second
return 59;

}
}
map<vector<int>, int> mp;
};

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