20170529-leetcode-582-Kill process

1.Description

Given n processes, each process has a unique PID (process id) and its PPID (parent process id).
Each process only has one parent process, but may have one or more children processes. This is just like a tree structure. Only one process has PPID that is 0, which means this process has no parent process. All the PIDs will be distinct positive integers.
We use two list of integers to represent a list of processes, where the first list contains PID for each process and the second list contains the corresponding PPID.
Now given the two lists, and a PID representing a process you want to kill, return a list of PIDs of processes that will be killed in the end. You should assume that when a process is killed, all its children processes will be killed. No order is required for the final answer.
Example 1:
Input:
pid = [1, 3, 10, 5]
ppid = [3, 0, 5, 3]
kill = 5
Output: [5,10]
Explanation:
3
/ \
1 5
/
10
Kill 5 will also kill 10.
Note:
The given kill id is guaranteed to be one of the given PIDs.
n >= 1.
解讀
給定一個進程列表，和該列表對應的父進程列表，輸入要殺掉的進程，輸出所有將要結束的進程

2.Solution

#代碼來自leecode，簡潔明瞭，下面做一下分析

class Solution(object):
    def killProcess(self, pid, ppid, kill):
        d = collections.defaultdict(list)
        #建立pid，ppid的對應表，如pid=[0,1,2,3,4,5],ppid=[0,1,1,1,1]
        #通過下面的代碼得到map{0：[0],1:[2,3,4,5]}
        for c, p in zip(pid, ppid): d[p].append(c)

        bfs = [kill]#最後的BFS結果中第一個元素是kill
        for i in bfs: bfs.extend(d.get(i, []))#把結果的子進程添加進來
        return bfs