690. Employee Importance（dfs bfs） 1834. Single-Threaded CPU

我做的是广度遍历，根据第一个员工找子员工放入队列中，再根据子员工找出其下一个员工。。。。

 public int getImportance(List employees, int id) {//存储员工idQueue queue=new ArrayDeque<>();queue.add(id);//员工id 价值Map map=new HashMap<>();//员工id 子员工Map> map1=new HashMap<>();for (int i = 0; i < employees.size(); i++) {map.put(employees.get(i).id,employees.get(i).importance);map1.put(employees.get(i).id,employees.get(i).subordinates);}int res=0;while (!queue.isEmpty()){res+=map.get(queue.peek());
//            增加子员工for (int i = 0; i < map1.get(queue.peek()).size(); i++) {queue.add(map1.get(queue.peek()).get(i));}queue.poll();}return res;}

官方给出深度遍历

class Solution {Map map = new HashMap();public int getImportance(List employees, int id) {for (Employee employee : employees) {map.put(employee.id, employee);}return dfs(id);}public int dfs(int id) {Employee employee = map.get(id);int total = employee.importance;List subordinates = employee.subordinates;for (int subId : subordinates) {total += dfs(subId);}return total;}
}作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/employee-importance/solution/yuan-gong-de-zhong-yao-xing-by-leetcode-h6xre/
来源：力扣（LeetCode）
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单线程cpu 短作业优先算法

class Solution {class Task{int id;int enqueueTime;int processingTime;public Task(int id, int enqueueTime, int processingTime){this.id = id;this.enqueueTime = enqueueTime;this.processingTime = processingTime;}}public int[] getOrder(int[][] tasks) {int len = tasks.length;List taskList = new ArrayList<>();for(int i=0; i t1.enqueueTime - t2.enqueueTime);//利用最小堆获取下个要执行的任务PriorityQueue minHeap = new PriorityQueue<>((t1,t2) -> {if(t1.processingTime == t2.processingTime){//当执行时间相同时，根据id升序return t1.id - t2.id;}else{//当执行时间不同时，根据执行时间升序return t1.processingTime - t2.processingTime;}});long now = 0;//当前时间，使用long防止int溢出int i = 0;//taskList的坐标int[] ret = new int[len];int p = 0;//ret的坐标while(i

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