Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU, find the exclusive time of these functions.

Each function has a unique id, start from 0 to n-1. A function may be called recursively or by another function.

A log is a string has this format : function_id:start_or_end:timestamp. For example, "0:start:0" means function 0 starts from the very beginning of time 0. "0:end:0" means function 0 ends to the very end of time 0.

Exclusive time of a function is defined as the time spent within this function, the time spent by calling other functions should not be considered as this function‘s exclusive time. You should return the exclusive time of each function sorted by their function id.

Example 1:

Input:
n = 2
logs = 
["0:start:0",
 "1:start:2",
 "1:end:5",
 "0:end:6"]
Output:[3, 4]
Explanation:
Function 0 starts at time 0, then it executes 2 units of time and reaches the end of time 1. 
Now function 0 calls function 1, function 1 starts at time 2, executes 4 units of time and end at time 5.
Function 0 is running again at time 6, and also end at the time 6, thus executes 1 unit of time. 
So function 0 totally execute 2 + 1 = 3 units of time, and function 1 totally execute 4 units of time.

Note:

1. Input logs will be sorted by timestamp, NOT log id.
2. Your output should be sorted by function id, which means the 0th element of your output corresponds to the exclusive time of function 0.
3. Two functions won‘t start or end at the same time.
4. Functions could be called recursively, and will always end.
5. 1 <= n <= 100

思路：

用一个数组记录当前id的时间，使用一个栈来保存当前的id，类似于后进先出，

如果当前id为end，那么栈顶一定也是相同的id，更新数组里面相应id的时间。

如果当前id为start，那么先更新上一个栈顶的id的时间，再将此id入栈。

最后保存当前的time，留作下一次更新。

vector<int> exclusiveTime(int n, vector<string>& logs) 
    {
        vector<int>ret(n), sta;
        int id, time, last;
        for (auto log:logs)
        {
            for (auto& c : log) if (c == ‘:‘)c = ‘ ‘;
            stringstream ss(log);            
            char s[9];
            ss >> id >> s >> time;
            //sscanf(log.c_str(), "%d:%[^:]:%d", &id, s, &time);
            //cout << id << " " << s << " " << time<<endl;
            if (s[0]==‘s‘)
            {
                if (sta.size() > 0)ret[sta.back()] += time - last;
                sta.push_back(id);                
            }
            else
            {
                ret[sta.back()] += ++time - last;
                sta.pop_back();
            }
            last = time;
        }
        return ret;
    }

[leetcode-636-Exclusive Time of Functions]