1116 · Exclusive Time of Functions
Algorithms
Medium
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.
Input logs will be sorted by timestamp, NOT log id.
Your output should be sorted by function id, which means the 0th element of your output corresponds to the exclusive time of function 0.
Two functions won’t start or end at the same time.
Functions could be called recursively, and will always end.
1 <= n <= 100
Example
Example 1:
Input:
2
0:start:0
1:start:2
1🔚5
0🔚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.
Example 2:
Input:
3
0:start:0
1:start:2
2:start:3
2🔚4
1🔚5
0🔚6
1:start:7
1🔚10
Output:
3 6 2
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 1 units of time.
Function 1 calls function 2, function 2 starts at time 3, executes 2 units of time.
Function 1 is running again at time 5, and also end at the time 5, thus executes 1 unit of time.
Function 0 is running again at time 6, and also end at the time 6, thus executes 1 unit of time.
Function 1 starts at time 7, and also end at the time 10, thus executes 4 unit of time.
So function 0 totally execute 2+1 = 3, units of time, and function 1 totally execute 1+1+4 = 6 units of time.function 2 totally execute 2 units of time.
解法1：
用stack。
每次都是stack.top()在运行。当有新来的任务时，我们要把stack.top()的时间累加
res[topNode.taskId] += startTime - topNode.startTime;
当新来的任务结束后，我们要把stack.top()的startTime改成当前时间+1。
struct Node {
int taskId;
int startTime;
Node(int tId = 0, int sT = 0) : taskId(tId), startTime(sT) {}
};
class Solution {
public:
/**
* @param n: a integer
* @param logs: a list of strings
* @return: return a list of integers
*/
vector<int> exclusiveTime(int n, vector<string> &logs) {
int logsLen = logs.size();
//map<int, int> mp; //<taskId, totalUsedTime>
vector<int> res(n, 0);
stack<Node> stk;
for (int i = 0; i < logsLen; i++) {
stringstream ss(logs[i]);
vector<string> buf;
string token;
while (getline(ss, token, ':')) {
buf.push_back(token);
}
if (buf[1] == "start") {
int taskId = stoi(buf[0]);
int startTime = stoi(buf[2]);
if (!stk.empty()) {
auto topNode = stk.top();
stk.pop();
res[topNode.taskId] += startTime - topNode.startTime;
stk.push(topNode);
}
stk.push(Node(taskId, startTime));
} else { // "end"
int taskId = stoi(buf[0]);
int endTime = stoi(buf[2]);
if (!stk.empty()) { //this check may be redundant
auto topNode = stk.top();
stk.pop();
int usedTime = endTime - topNode.startTime + 1;
res[taskId] += usedTime;
if (!stk.empty()) {
topNode = stk.top();
stk.pop();
topNode.startTime = endTime + 1;
stk.push(topNode);
}
}
}
}
return res;
}
};