582. Kill Process

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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:

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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.

解法1：

先把问题条件转化为一个图，然后对图做一个BFS就可以了。
C++

1

Java

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public class Solution {
    public List<Integer> killProcess(List<Integer> pid, List<Integer> ppid, int kill) {
        List<Integer> res = new ArrayList<Integer>();
        if (pid == null || ppid == null) {
            return res;
        }
        if (pid.size() != ppid.size()) {
            return res;
        }
        // Convert to a graph
        HashMap<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();
        for (int i = 0; i < pid.size(); i++) {
            int child  = pid.get(i);
            int parent = ppid.get(i);
            if (!map.containsKey(parent)) {
                List<Integer> temp = new ArrayList<Integer>();
                temp.add(child);
                map.put(parent, temp);
            } else {
                map.get(parent).add(child);
            }
        }
        // do a bfs on the graph to get all childs
        Queue<Integer> queue = new LinkedList<Integer>();
        queue.offer(kill);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int id = queue.poll();
                res.add(id);
                List<Integer> children = map.get(id);
                if (children != null) { // important! check if there exists children
                    for (int c : children) {
                        queue.offer(c);
                    }
                }
            }
        }
        return res;        
    }
}