636. Exclusive Time of Functions ---Medium !!!!!! 自己对定义理解不清楚

 On a single-threaded CPU, we execute a program containing n functions. Each function has a unique ID between 0 and n-1.

Function calls are stored in a call stack: when a function call starts, its ID is pushed onto the stack, and when a function call ends, its ID is popped off the stack. The function whose ID is at the top of the stack is the current function being executed. Each time a function starts or ends, we write a log with the ID, whether it started or ended, and the timestamp.

You are given a list logs, where logs[i] represents the ith log message formatted as a string "{function_id}:{"start" | "end"}:{timestamp}". For example, "0:start:3" means a function call with function ID 0 started at the beginning of timestamp 3, and "1:end:2" means a function call with function ID 1 ended at the end of timestamp 2. Note that a function can be called multiple times, possibly recursively.

A function's exclusive time is the sum of execution times for all function calls in the program. For example, if a function is called twice, one call executing for 2 time units and another call executing for 1 time unit, the exclusive time is 2 + 1 = 3.

Return the exclusive time of each function in an array, where the value at the ith index represents the exclusive time for the function with ID i.

 

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 the beginning of time 0, then it executes 2 for units of time and reaches the end of time 1.
Function 1 starts at the beginning of time 2, executes for 4 units of time, and ends at the end of time 5.
Function 0 resumes execution at the beginning of time 6 and executes for 1 unit of time.
So function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing.

Example 2:

Input: n = 1, logs = ["0:start:0","0:start:2","0:end:5","0:start:6","0:end:6","0:end:7"]
Output: [8]
Explanation:
Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself.
Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time.
Function 0 (initial call) resumes execution then immediately calls itself again.
Function 0 (2nd recursive call) starts at the beginning of time 6 and executes for 1 unit of time.
Function 0 (initial call) resumes execution at the beginning of time 7 and executes for 1 unit of time.
So function 0 spends 2 + 4 + 1 + 1 = 8 units of total time executing.

Example 3:

Input: n = 2, logs = ["0:start:0","0:start:2","0:end:5","1:start:6","1:end:6","0:end:7"]
Output: [7,1]
Explanation:
Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself.
Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time.
Function 0 (initial call) resumes execution then immediately calls function 1.
Function 1 starts at the beginning of time 6, executes 1 unit of time, and ends at the end of time 6.
Function 0 resumes execution at the beginning of time 6 and executes for 2 units of time.
So function 0 spends 2 + 4 + 1 = 7 units of total time executing, and function 1 spends 1 unit of total time executing.

 

Constraints:

• 1 <= n <= 100
• 1 <= logs.length <= 500
• 0 <= function_id < n
• 0 <= timestamp <= 109
• No two start events will happen at the same timestamp.
• No two end events will happen at the same timestamp.
• Each function has an "end" log for each "start" log.

A:

下面是自己的理解：但是是错误的。  原因在代码下方

class Solution {

public:

vector<int> exclusiveTime(int n, vector<string>& logs) {

vector<int> V(n,0);

// [pre, cur]=[start ,start], we can calculate first,

// [start, end] // its a pair ,thus we calculate end

// [end, end] // we need calculate end

// [end, start] we calculate nothing. NOTE: there might be some gap in between,

Event pre = {1,false, 0}, cur = {0, false,0};// does not matter what the initial value

for(int i =0; i< logs.size(); i++){

pre = cur;

helper(logs[i], cur);

if(i != 0){

if(!cur.isStart){// it is end

V[cur.id] += cur.time - pre.time + (pre.isStart? 1: 0) ;

}else if(pre.isStart){ //now cur is start, [start ,start], calculate first

V[pre.id] = cur.time - pre.time;

}

}

}

return V;

}

private:

struct Event {

int id;

bool isStart;

int time;

};

void helper(const string &log, Event &evt){

int posColon1 = log.find(':', 0);

evt.id = stoi(log.substr(0,posColon1));

evt.isStart = log[posColon1+1] == 's';

int posColon2 = log.find(':', posColon1+1);

evt.time = stoi(log.substr(posColon2+1, log.size() - posColon2-1));

}

};

原因是， 如果stack里有，  那么会继续执行，  而中途又被打断。

所以，不能只看最近2个的情况。

class Solution {

public:

vector<int> exclusiveTime(int n, vector<string>& logs) {

// [pre, cur]=[start ,start], we can calculate first,

// [start, end] // its a pair ,thus we calculate end

// [end, end] // we need calculate end

// [end, start] we calculate nothing. NOTE: there might be some gap in

// between,

vector<int> V(n, 0);

stack<Event> S; // save only the ones that is running on stack

for (int i = 0; i < logs.size(); i++) {

Event cur = helper(logs[i]);

if (!cur.isStart) {

V[cur.id] += cur.time - S.top().time + 1;

S.pop();

if (!S.empty()) {

S.top().time = cur.time + 1; // task on top will resume

}

} else { // cur.isStart == true

if (!S.empty() && S.top().isStart) {

V[S.top().id] += cur.time - S.top().time;

}

S.push(cur);

}

}

return V;

}

private:

struct Event {

int id;

bool isStart;

int time;

};

Event helper(const string& log) {

int posColon1 = log.find(':', 0);

int id = stoi(log.substr(0, posColon1));

bool isStart = log[posColon1 + 1] == 's';

int posColon2 = log.find(':', posColon1 + 1);

int time = stoi(log.substr(posColon2 + 1, log.size() - posColon2 - 1));

Event evt = {id, isStart, time};

return evt;

}

};