636. Exclusive Time of Functions

最新推荐文章于 2020-10-06 20:42:22 发布

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博客围绕单线程CPU执行函数展开，介绍了函数日志格式，如“{function_id}:{“start” | “end”}:{timestamp}”，并定义了函数的独占时间。通过示例展示如何根据输入的日志计算每个函数的独占时间，还提到单CPU执行函数的规则。

题目

  计算函数执行时长。

On a single threaded CPU, we execute some functions. Each function has a unique id between 0 and N-1.

We store logs in timestamp order that describe when a function is entered or exited.

Each log is a string with this format: “{function_id}:{“start” | “end”}:{timestamp}”. For example, “0:start:3” means the function with id 0 started at the beginning of timestamp 3. “1: end :2” means the function with id 1 ended at the end of timestamp 2.

A function’s exclusive time is the number of units of time spent in this function. Note that this does not include any recursive calls to child functions.

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 the beginning of time 0, then it executes 2 units of time and reaches the end of time 1.
Now function 1 starts at the beginning of time 2, executes 4 units of time and ends at time 5.
Function 0 is running again at the beginning of time 6, and also ends at the end of time 6, thus executing 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.

代码
单cpu执行某个函数，只能执行完了再返回上层函数。所以不会出现 0 start 1 start 0 end 1 end 这种测试样例。

class Solution {
   public int[] exclusiveTime(int n, List<String> logs) {
       
        int[] res = new int[n];
        Stack<Integer> stack = new Stack<>();
        int prevTime = 0;
       
        for (String log : logs) {
            
            String[] parts = log.split(":");
            
            if (!stack.isEmpty()) 
                res[stack.peek()] +=  Integer.parseInt(parts[2]) - prevTime; 
            
            prevTime = Integer.parseInt(parts[2]);
            
            if (parts[1].equals("start")) 
                stack.push(Integer.parseInt(parts[0]));
            else {
                res[stack.pop()]++;//记录时间段的end,即下一个时间段的begin
                prevTime++;
            }
        }
       
        return res;
    }
}