LeetCode #582 - Kill Process

最新推荐文章于 2020-05-25 09:00:44 发布

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这是一个关于计算机进程的问题，给定一个包含进程ID（PID）和父进程ID（PPID）的树状结构。当你要杀死一个具有特定PID的进程时，所有其子进程也将被杀死。你需要返回最终会被杀死的所有进程的PID列表。

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 = 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.

class Solution {
public:
    vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
        unordered_map<int,vector<int>> hash; //构建父进程到子进程的映射
        for(int i=0;i<ppid.size();i++)
        {
            if(ppid[i]==0) continue;
            if(hash.count(ppid[i])==0) hash[ppid[i]]={pid[i]};
            else hash[ppid[i]].push_back(pid[i]);
        }
        
        vector<int> result;
        queue<int> q; // 利用队列，从某个进程开始，遍历所有的子进程
        q.push(kill);
        while(!q.empty())
        {
            int id=q.front();
            q.pop();
            result.push_back(id);
            for(int x:hash[id]) q.push(x);
        }
        return result;
    }
};