[LeetCode] 582. Kill Process 终止进程

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.

给两个数组，一个是进程，一个是进程数组中的每个进程的父进程组成的数组。结束了某一个进程，其所有的子进程都需要结束，由于一个进程可能有多个子进程，所以要理清父子进程的关系。

解法：使用一个哈希表，建立进程和其所有子进程之间的映射。先把把要结束的进程放入结果中，然后将所有子进程以及以下的进程放入结果中。

Python: DFS, Time: O(n), Space: O(n)

class Solution(object):
    def killProcess(self, pid, ppid, kill):
        """
        :type pid: List[int]
        :type ppid: List[int]
        :type kill: int
        :rtype: List[int]
        """
        def killAll(pid, children, killed):
            killed.append(pid)
            for child in children[pid]:
                killAll(child, children, killed)

        result = []
        children = collections.defaultdict(set)
        for i in xrange(len(pid)):
            children[ppid[i]].add(pid[i])
        killAll(kill, children, result)
        return result

Python: BFS, Time: O(n), Space: O(n)

class Solution(object):
    def killProcess(self, pid, ppid, kill):
        """
        :type pid: List[int]
        :type ppid: List[int]
        :type kill: int
        :rtype: List[int]
        """
        result = []
        children = collections.defaultdict(set)
        for i in xrange(len(pid)):
            children[ppid[i]].add(pid[i])
        q = collections.deque()
        q.append(kill)
        while q:
            p = q.popleft()
            result.append(p)
            for child in children[p]:
                q.append(child)
        return result　　

C++:

class Solution {
public:
    vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
        vector<int> res;
        queue<int> q{{kill}};
        unordered_map<int, vector<int>> m;
        for (int i = 0; i < pid.size(); ++i) {
            m[ppid[i]].push_back(pid[i]);
        }
        while (!q.empty()) {
            int t = q.front(); q.pop();
            res.push_back(t);
            for (int p : m[t]) {
                q.push(p);
            }
        }
        return res;
    }
};

C++:

class Solution {
public:
    vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
        vector<int> res;
        unordered_map<int, vector<int>> m;
        for (int i = 0; i < pid.size(); ++i) {
            m[ppid[i]].push_back(pid[i]);
        }
        helper(kill, m, res);
        return res;
    }
    void helper(int kill, unordered_map<int, vector<int>>& m, vector<int>& res) {
        res.push_back(kill);
        for (int p : m[kill]) {
            helper(p, m, res);
        }
    }
};

原文地址：https://www.cnblogs.com/lightwindy/p/9532814.html