[leetcode] 636. Exclusive Time of Functions

Description

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?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

分析

题目的意思是：给出一些函数的执行log，算出每个函数的执行时间。

• 这一道题很明显就是要对字符串进行处理，要用到c++的find,find_last_of,substr字符串处理函数。
• 核心就是按照":"对字符串进行分割，转化为运算，用栈来辅助解决运算问题。注意比如如果一个函数从6开始，又从6结束，则还算是执行了一个时间单元。这也是代码中++res[t]的原因。
• 如果是start，则压入栈中，如果是end，则执行出栈操作，因为是单CPU，所以我们在出栈或者开启另一个函数之前，都需要计算上一个函数的运行时间。

代码

class Solution {
public:
    vector exclusiveTime(int n, vector& logs) {
        vector res(n,0);
        stack st;
        int preTime=0;
        for(string log:logs){
            int i1=log.find(":");
            int i2=log.find_last_of(":");
            int idx=stoi(log.substr(0,i1));
            string type=log.substr(i1+1,i2-i1-1);
            int time=stoi(log.substr(i2+1));
            if(!st.empty()){
                res[st.top()]+=time-preTime;
            }
            preTime=time;
            if(type=="start"){
                st.push(idx);
            }else{
                auto t=st.top();
                st.pop();
                ++res[t];
                ++preTime;
            }
        }
        return res;
    }
};

参考文献

[LeetCode] Exclusive Time of Functions 函数的独家时间