Leetcode one question per day (636. exclusive time of functions)

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 “12” 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”,“15”,“06”]
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”,“05”,“0:start:6”,“06”,“07”]
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”,“05”,“1:start:6”,“16”,“07”]
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.

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Create a stack to hold all start, Every time I meet one start Then settle the previous start To the current start Running time of , This time is added to the previous start Of function On the running time , Because from this moment on , Previous start Of function It's no longer implemented . If you encounter end Then directly settle the current function Running time of ( At present end Time - To the top of the stack start Time ), After settlement, you need to modify the previous start Time for , Because from now on end From the time of , Previous start Of is running again . Check that the stack is empty , also , The premise of all this is based on logs It is recorded according to the time axis

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impl Solution {
    
    pub fn exclusive_time(n: i32, logs: Vec<String>) -> Vec<i32> {
    
        let mut ans = vec![0; n as usize];
        let mut stack: Vec<(i32, i32)> = Vec::new();
        for log in logs {
    
            let group: Vec<String> = log.split(":").map(|v| v.to_owned()).collect();
            let num = group[0].parse::<i32>().unwrap();
            let typ = group[1].clone();
            let t = group[2].parse::<i32>().unwrap();
            if typ == "start" {
    
                if let Some(prev) = stack.pop() {
    
                    ans[prev.0 as usize] += t - prev.1;
                    stack.push(prev);
                }
                stack.push((num, t));
                continue;
            }
            let prev = stack.pop().unwrap();
            ans[prev.0 as usize] += t - prev.1 + 1;
            if let Some(mut p) = stack.pop() {
    
                p.1 = t + 1;
                stack.push(p);
            }
        }
        ans
    }
}