zoukankan      html  css  js  c++  java
  • Course Schedule II

    Course Schedule II

    问题:

    There are a total of n courses you have to take, labeled from 0 to n - 1.

    Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

    Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

    There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

    思路:

      dfs

    我的代码:

    public class Solution {
         public int[] findOrder(int numCourses, int[][] prerequisites) {
            if(prerequisites==null || prerequisites.length==0) {
                int[] a = new int[numCourses];
                for(int i=0;i<numCourses;i++) {
                    a[i]=i;
                }
                return a;
            }
            HashSet<Integer>[] graph = new HashSet[numCourses];
            for(int i=0; i<numCourses; i++)
                graph[i] = new HashSet<Integer>();
            boolean[] visited = new boolean[numCourses];
            boolean[] visiting = new boolean[numCourses];
            
            for(int i=0; i<prerequisites.length; i++)
            {
                graph[prerequisites[i][1]].add(prerequisites[i][0]);
            }
            for(int i=0; i<numCourses; i++)
            {
                if(visited[i]) continue;
                if(!helper(graph, visited, visiting, i)) return new int[0];
            }
            int[] nums = new int[numCourses];
            int i = 0;
            for(int count=rst.size()-1; count>=0; count--)
                nums[i++] = rst.get(count);
            return nums;
        }
        private List<Integer> rst = new ArrayList<Integer>();
        public boolean helper(HashSet<Integer>[] graph, boolean[] visited, boolean[] visiting, int cur)
        {
            if(visiting[cur]) return false;
            visiting[cur] = true;
            for(Integer neighbor: graph[cur])
            {
                if(visited[neighbor]) continue;
                if(!helper(graph, visited, visiting, neighbor)) return false;  
            }
            visited[cur] = true;
            visiting[cur] = false;
            rst.add(cur);
            return true;
        }
    }
    View Code

    学习之处:

    • 什么叫一个节点访问结束了,只有当这个节点的所有邻居都访问完了,才算结束了,此时visited[cur] = true;
  • 相关阅读:
    Appium Android sdk自动化工具安装
    roboframework-ride运行案例时报 Error 267 错误问题
    Flask 编写http接口api及接口自动化测试
    ssh免密码登录快速配置方法
    Ansible 介绍和使用
    Python 递归返回树形菜单JSON串 <flask>
    python学习之路web框架续
    python学习之路web框架续
    python学习之路web框架
    python学习之路前端-Dom
  • 原文地址:https://www.cnblogs.com/sunshisonghit/p/4511979.html
Copyright © 2011-2022 走看看