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.
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| 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.
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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.
Solution#
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| class Solution {
/*
["0:start:0","0:start:2","0:end:5","0:start:6","0:end:6","0:end:7"]
0-----| ---|
0 |--------|
0 |--|
0--1--2--3--4--5--6--7--8
2
["0:start:0","1:start:2","1:end:5","0:end:6"]
0-----| |--|
1 |--------|
0--1--2--3--4--5--6--7--8
*/
public int[] exclusiveTime(int n, List<String> logs) {
int[] res = new int[n];
Deque<Integer> stack = new ArrayDeque<>();
for (int i = 0; i < logs.size(); i++) {
String[] log = logs.get(i).split(":");
int id = Integer.parseInt(log[0]);
int time = Integer.parseInt(log[2]);
if ("start".equals(log[1])) {
stack.push(i);
} else {
String[] prevLog = logs.get(stack.pop()).split(":");
int prevId = Integer.parseInt(prevLog[0]);
int prevTime = Integer.parseInt(prevLog[2]);
int amount = time - prevTime + 1;
res[id]+= amount;
if (stack.size() > 0) {
int topId = Integer.parseInt(logs.get(stack.peek()).split(":")[0]);
res[topId] -= amount;
}
}
// System.out.println("s = " + stack + " res = " + Arrays.toString(res));
}
return res;
}
}
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