Leetcode:Kill Process
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题目
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 / 10
Kill 5 will also kill 10.
思路
本题刚开始时,采用递归的思路,找到父进程的一个子进程后,进行递归,以此进行,但是超时。可以先将父进程和子进程之间建立一个HashMap,在查找时的复杂度为O(1),就不会超时了。
代码
public class Solution { public List<Integer> killProcess(List<Integer> pid, List<Integer> ppid, int kill) { HashMap<Integer,List<Integer>> hashProcess = new HashMap<Integer,List<Integer>>(); for(int i=0;i<ppid.size();i++){ if(hashProcess.containsKey(ppid.get(i))) hashProcess.get(ppid.get(i)).add(pid.get(i)); else{ List<Integer> listProcess = new ArrayList<Integer>(); listProcess.add(pid.get(i)); hashProcess.put(ppid.get(i),listProcess); } } List<Integer> killedList = new ArrayList<Integer>(); killProcessAll(hashProcess,kill,killedList); return killedList; } public void killProcessAll(HashMap<Integer,List<Integer>> pidMap, int kill, List<Integer> proKilledList){ proKilledList.add(kill); List<Integer> pidList = pidMap.get(kill); if(pidList == null) //如果不进行判断的话,pidList.size()会报空指针异常 return; for(int i=0;i<pidList.size();i++) killProcessAll(pidMap,pidList.get(i),proKilledList); }}
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