LeetCode #636 - Exclusive Time of Functions-优快云博客

本文探讨了在非抢占式的单线程CPU环境下，如何计算n个函数的独占执行时间。通过解析运行日志，文章详细介绍了如何区分函数间的递归调用和直接调用，从而准确计算每个函数的实际执行时间，排除被调用函数的时间开销。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

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

class Solution {
public:
    vector<int> exclusiveTime(int n, vector<string>& logs) {
        vector<int> result(n,0);
        stack<int> s; // 进程开始时压栈，进程结束时出栈
        int pre_time=0;
        string pre_state;
        for(int i=0;i<logs.size();i++)
        {
            vector<string> v(3);
            istringstream iss(logs[i]);
            getline(iss,v[0],':');
            getline(iss,v[1],':'); 
            getline(iss,v[2],':');
            int cur=get_num(v[0]);
            int cur_time=get_num(v[2]);
            if(i==0) 
            {
                pre_time=cur_time;
                pre_state=v[1];
                s.push(cur);
            }
            else
            {
                if(v[1]=="start")
                {
                    if(pre_state=="start") result[s.top()]+=cur_time-pre_time;
                    else result[s.top()]+=cur_time-pre_time-1;
                    s.push(cur);
                }
                else if(v[1]=="end")
                {
                    if(pre_state=="start") result[s.top()]+=cur_time-pre_time+1;
                    else result[s.top()]+=cur_time-pre_time;
                    s.pop();
                }
                pre_time=cur_time;
                pre_state=v[1];
            }
        }
        return result;
    }
    
    int get_num(string s)
    {
        int ans=0;
        for(int i=0;i<s.size();i++)
        {
            ans*=10;
            ans+=s[i]-'0';
        }
        return ans;
    }
};