LeetCode Kill Process

原题链接在这里：https://leetcode.com/problems/kill-process/description/

题目：

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:

1. The given kill id is guaranteed to be one of the given PIDs.
2. n >= 1.

题解：

通过存pid 和 ppid 的两个list 建立起 parent 和 它对应children list的map.

然后用DFS 或 BFS 从给出的process id走起就好.

Time Complexity: O(n). n = pid.size().

Space: O(n).

AC Java:

 class Solution {
     public List<Integer> killProcess(List<Integer> pid, List<Integer> ppid, int kill) {
         HashMap<Integer, List<Integer>> hm = new HashMap<Integer, List<Integer>>();
         for(int i = 0; i<ppid.size(); i++){
             int parent = ppid.get(i);
             if(parent > 0){
                 List<Integer> item = hm.getOrDefault(parent, new ArrayList<Integer>());
                 item.add(pid.get(i));
                 hm.put(parent, item);
             }
         }

         List<Integer> res = new ArrayList<Integer>();
         LinkedList<Integer> que = new LinkedList<Integer>();
         que.add(kill);
         while(!que.isEmpty()){
             int cur = que.poll();
             res.add(cur);
             if(hm.containsKey(cur)){
                 que.addAll(hm.get(cur));
             }
         }

         return res;
     }
 }