第十三周：( LeetCode582) Kill Process（c++）

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

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.

思路：本题其实就是输出某棵子树的所有子孙。求解过程中主要用了map的数据结构。因为map的查找是基于哈希的查找，所以效率没得说。如果直接查找会超时！

代码：

class Solution {
public:
    vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
        int nums=pid.size();
        int kill_index=0;
        vector<int> to_kill;
        //int存储ppid，vector<int>存储该ppid对应的孩子
        map<int,vector<int>> node;
        for(int i=0;i<nums;i++)
            node[ppid[i]].push_back(pid[i]);
        to_kill.push_back(kill);
        //其实遍历的过程就是BFS,本题把vector当queue用，因为函数要返回vector<int>的类型
        while(kill_index<to_kill.size()){
            kill=to_kill[kill_index];
            //通过node[kill]查找的效率比较高
            for(int i=0;i<node[kill].size();i++)
                to_kill.push_back(node[kill][i]);
            kill_index++;
        }
        return to_kill;
    }
};