[Leetcode] 636. Exclusive Time of Functions 解题报告

题目：

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：

一道考查栈这种数据结构的题目。我们顺序遍历logs：如果当前log记录的是某个任务的开始时间，那么既有可能是开始了一个新的任务，也有可能是当前的任务取代了上一个任务，此时我们就需要计算上一个任务（如果有的话）在这个任务开始之前执行的时间。最后把这个任务的id也压入堆栈中。如果当前log记录的是某个任务的结束时间，那么就计算该任务在这段时间内的执行时间，完了将其出栈即可。

代码：

class Solution {
public:
    vector exclusiveTime(int n, vector& logs) {
        vector ret(n, 0);
        stack st;  // record the id of the last task
        int id, current_time, last_time = 0;
        bool start;
        for (int i = 0; i < logs.size(); ++i) {
            getItems(logs[i], id, start, current_time);
            if (start) {
                if (!st.empty()) {
                    int last = st.top();
                    ret[last] += (current_time - last_time);
                }
                st.push(id);
                last_time = current_time;
            }
            else {
                int last = st.top();
                ret[last] += (current_time + 1 - last_time);
                st.pop();
                last_time = current_time + 1;
            }
        }
        return ret;
    }
private:
    void getItems(const string &s, int &id, bool &start, int &time) {
        int first = s.find_first_of(':');
        int last = s.find_last_of(':');
        id = stoi(s.substr(0, first));
        if (s.substr(first, last - first + 1) == ":start:") {
            start = true;
        }
        else {
            start = false;
        }
        time = stoi(s.substr(last + 1));
    }
};