Exclusive Time of Functions问题及解法

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.

示例:

Input:n = 2logs = ["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.

问题分析：

利用栈结构，记录好一个程序段整体运行时间和程序其中其他程序已使用的时间片段，不同的程序运行时间分别累计相加，最终可得答案。

过程详见代码：

class Solution {public:    vector<int> exclusiveTime(int n, vector<string>& logs) {vector<int> res(n, 0);int id, timepoint;stack<pair<int, int>> s;string flag;unordered_map<int, int> num;num[0] = 0;for (int i = 0; i < logs.size(); i++){analysize(logs[i], id, flag, timepoint);if (flag == "start"){s.push(pair<int, int>(id, timepoint));num[s.size()] = 0;}else{pair<int, int> p = s.top();res[id] += timepoint - p.second + 1 - num[s.size()];s.pop();num[s.size()] += timepoint - p.second + 1;}}return res;}void analysize(string s, int& id, string& flag, int& timepoint){stringstream ss(s);string str;getline(ss, str, ':');id = stoi(str);getline(ss, str, ':');flag = str;getline(ss, str, ':');timepoint = stoi(str);}};