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

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 

思路

這題最難理解的地方是

the very end of 5  =  the very beginning of 6 

弄清這點，程式碼就好理解了

程式碼

 1 class Solution {
 2     public int[] exclusiveTime(int n, List<String> logs) {
 3         int[] res = new int[n];
 4         Stack<Integer> stack = new Stack();
 5         int pre = 0;
 6         for (String log : logs) {
 7             String[] arr = log.split(":");
 8             // function_id:start_or_end:timestamp
 9             if (arr[1].equals("start")) {
10                 if (!stack.isEmpty()) {
11                     res[stack.peek()] += Integer.parseInt(arr[2]) - pre;
12                 }
13                 stack.push(Integer.parseInt(arr[0]));
14                 pre = Integer.parseInt(arr[2]);
15             } else {
16                 res[stack.pop()] += Integer.parseInt(arr[2]) - pre + 1;
17                 pre = Integer.parseInt(arr[2]) + 1;
18             }
19         }
20         return res;
21     }
22 }