636. Exclusive Time of Functions

Problem:

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.

Analysis:
用stack来保存log的index。
1. 如果遇到start而且stack不为空，则更新stack.peek()。说明保存在stack.peek()的log暂停，开始新的log。 然后把当前log的index加入stack里面
2. 如果遇到end, 更新stack.pop()。当前log停止了。如果是在5 end，由于end是在末尾，所以需要加1。

Solution:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

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