彭博面试题：日程表 Busy View 合并重叠时间段

[QABANK] Calendar Busy View

Problem Description:
Summary: Given list of events of a given day in a calendar, write an algorithm to return a list 
of busy view time slots. Busy view is created by consolidating adjacent and overlapping event 
time slots without showing details of individual events.

Details: Each event in calendar has a start time, end time & some title. Events can start at 
any minute (granularity at the minute only, no seconds).

Sample inputs - Expected outputs

Input: list of following events

    (100,300,"Some Event")     // 1:00 am to 3:00 am
    (115,145,"Some Event")
    (145,215,"Some Event")
    (200,400,"Some Event")
    (215,230,"Some Event")
    (215,415,"Some Event")
    (600,700,"Some Event")
    (500,600,"Some Event")

Output: Based on above events, my busy view should show like this:

    (100,415)   // Busy from 1am to 4:15am
    (500,700)   // Busy from 5am to 7am

这题要求把当天的所有日程事件按照时间合并成 Busy View，即把所有重叠或相邻的时间段合并成连续区间。典型解法是先按开始时间排序，再遍历并逐段合并。最终输出就是若干个不重叠的忙碌时间块。

彭博面试题：识别最先崩溃的进程（树结构中的根故障）

An operating system has the following rules:

1. Each process is spawned from a parent process except the first process.
2. The parent process is aware of its children.
3. If a parent process is aborted, all the child processes get aborted.

One of the processes crashed and brought down a lot of other processes along with it.
A log parser found the set of all the processes which have crashed. This set of processes,
along with sub-process child relationships for each process, is then passed to your function.

Identify the id of the process which crashed first.

Example tree:

        15
       /  \
      7    5
     / \    \
    9   4    1
   / \
  2  11

Input:
[
    Process(9, [2,11]),
    Process(11, []),
    Process(4, []),
    Process(7, [9,4]),
    Process(2, []),
]

Output: 7

题目给出所有已崩溃的进程及其子进程关系。由于父进程崩溃会导致所有后代都崩溃，所以最先崩溃的一定是这些崩溃节点中的“最高祖先”，也就是那个没有崩溃父节点的进程。解法是把所有 crashed pid 存入集合，再查每个节点是否有 crashed 父节点，找出那个没有父节点的即可。