LeetCode Exclusive Time of Functions

· 513 words · 2 minute read
2017年7月
LeetCode

636. Exclusive Time of Functions

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:end: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.

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 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.

Note:

  1. Input logs will be sorted by timestamp, NOT log id.
  2. Your output should be sorted by function id, which means the 0th element of your output corresponds to the exclusive time of function 0.
  3. Two functions won’t start or end at the same time.
  4. Functions could be called recursively, and will always end. 5.1 <= n <= 100

方法

这道题实际上个人觉得值得细细品味。刷题到现在看到这种题能知道用Stack值得欣慰，但是怎么存储每个函数的使用时间又让我有点蒙圈。实际上这道题不仅包含了函数先进后出的思想，在计算函数的使用时间时又使用了先进先出的思想。

用Stack存储函数的执行顺序，用res计算在下一个函数执行之前函数已经执行的时间。

public class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        int[] res = new int[n];
        Stack<Integer> stack = new Stack<Integer>();
        String[] s = logs.get(0).split(":");
        stack.push(Integer.parseInt(s[0]));
        int pre = Integer.parseInt(s[2]);
        for(int i = 1; i < logs.size(); i++){
            String[] str = logs.get(i).split(":");
            if(str[1].equals("start")){
                if(!stack.isEmpty())
                    res[stack.peek()] += Integer.parseInt(str[2]) - pre;
                stack.push(Integer.parseInt(str[0]));
                pre = Integer.parseInt(str[2]);
            }else{
                res[stack.peek()] += Integer.parseInt(str[2]) - pre + 1;
                stack.pop();
                pre = Integer.parseInt(str[2]) + 1;
            }
        }
        return res;
    }
}
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