Kill Process

题目描述：

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 = 5Output: [5,10]Explanation:            3         /   \        1     5             /            10Kill 5 will also kill 10.

Note:

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

思路：

第一步，建立一个哈希表，哈希表的关键字是ppid里的元素，哈希表的值是以ppid中元素为父线程的线程。

第二步，建立一个队列，队列中存放将要kill的线程，假设pp是队列中的首元素，那么将其弹出，加入到最终结果中，

              并将以pp为父线程的线程加入队列，直到队列为空。

代码：

class Solution {
public:
    vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
        vector<int> res;
        deque<int> killing;
        unordered_map<int,vector<int>> hash;
        for (int i=0; i<ppid.size(); ++i)
            hash[ppid[i]].push_back(pid[i]);
        
        killing.push_back(kill);
        while(!killing.empty())
        {
            int pp=killing.front();
            res.push_back(pp);
            killing.pop_front();
            for (int i=0; i<hash[pp].size(); ++i)
                killing.push_back(hash[pp][i]);
        }
        return res;
    }
    
};