636. Exclusive Time of Functions

问题描述：

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

 

解题思路：

主要思路：

　　要记录上一个log的相关参数，与当前log相比较，在这里我的关注重点为可能的出现的状态：

　　　　a. start start: 可能出现在前后两个不同的任务中，我们应当把时间差加到前面一个的任务中 

　　　　b. start end：出现在同一个任务中，时间差计算为：end - start + 1，因为还要算上start的1个时间单位

　　　　c. end end：可能出现在两个不同的任务中，应当把时间差加到后面一个任务中

　　　　d. end start： 可能出现在两个不同任务中，需要注意的是，此时时间差不一定为1，那么这段时间应该是最近一个还没有结束的任务在占用。

为了保证d的情况，我们用栈来记录当前正在运行的任务。

需要注意的是：

一定要保证出栈入栈成双成对！

 

代码：

class Solution {
public:
    vector<int> exclusiveTime(int n, vector<string>& logs) {
        vector<int> ret(n, 0);
        if(logs.empty())
            return ret;
        stack<int> stk;
        int prev_id = -1;
        string prev_state = "start";
        int prev_time = 0;
        for(string s : logs){
            stringstream ss(s);
            string temp;
            getline(ss, temp, ':');
            int id = stoi(temp);
            getline(ss, temp, ':');
            string cur_state = temp;
            getline(ss, temp, ':');
            int t = stoi(temp);
            if(prev_id != -1){
                if(cur_state != prev_state){
                    if(cur_state == "end"){
                        ret[id] += (t - prev_time + 1);
                        stk.pop();
                    }else{
                        int time_diff = t - prev_time;
                        if(time_diff > 1 && !stk.empty()){
                            ret[stk.top()] += time_diff - 1;
                        }
                        stk.push(id);
                    }
                } else{
                    if(cur_state == "start"){
                        ret[prev_id] += t - prev_time;
                        stk.push(id);
                    }
                    else{
                        ret[id] += t - prev_time;
                        stk.pop();
                    }
                }
            }else{
                stk.push(id);
            }
            prev_state = cur_state;
            prev_id = id;
            prev_time = t;
        }
        return ret;
    }
};