• 210. Course Schedule II


    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.

    Example 1:

    Input: 2, [[1,0]] 
    Output: [0,1]
    Explanation: There are a total of 2 courses to take. To take course 1 you should have finished   
                 course 0. So the correct course order is [0,1] .

    Example 2:

    Input: 4, [[1,0],[2,0],[3,1],[3,2]]
    Output: [0,1,2,3] or [0,2,1,3]
    Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both     
                 courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. 
                 So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] .

    Note:

    1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
    2. You may assume that there are no duplicate edges in the input prerequisites.

    同207 Course Schedule,topological sort + BFS,https://www.cnblogs.com/fatttcat/p/10021339.html

    多一个额外步骤记下每次从queue中poll中的元素并加入result中

    时间:O(VE),空间:O(V)

    class Solution {
        public int[] findOrder(int numCourses, int[][] prerequisites) {
            int[] res = new int[numCourses];
            if(numCourses <= 0) return res;
            
            ArrayList[] graph = new ArrayList[numCourses];
            for(int i = 0; i < numCourses; i++)
                graph[i] = new ArrayList();
            
            int[] indegree = new int[numCourses];
            for(int[] pair : prerequisites) {
                graph[pair[1]].add(pair[0]);
                indegree[pair[0]]++;
            }
            
            Queue<Integer> q = new LinkedList<>();
            for(int i = 0; i < indegree.length; i++) {
                if(indegree[i] == 0)
                    q.offer(i);
            }
            
            int cnt = 0;
            while(!q.isEmpty()) {
                int cur = q.poll();
                res[cnt++] = cur;
                ArrayList tmp = graph[cur];
                for(int i = 0; i < tmp.size(); i++) {
                    --indegree[(int)tmp.get(i)];
                    if(indegree[(int)tmp.get(i)] == 0)
                        q.offer((int)tmp.get(i));
                }
            }
            
            for (int i = 0; i < numCourses; ++i) {
                if (indegree[i] != 0)
                    return new int[0];
            }
            return res;
        }
    }
  • 相关阅读:
    英语语法基础知识总结名词
    英语语法基础知识代词
    idea破解第二部
    Linux(centos)系统下安装fastdfs安装部署步骤
    死磕Synchronized底层实现概论偏向锁轻量级锁重量级锁
    深入理解Mysql事务隔离级别与锁机制
    事务的传播机制
    ICCV2021 | TokenstoToken ViT:在ImageNet上从零训练Vision Transformer
    计算机视觉CV技术指南文章汇总
    资源分享 | PyTea:不用运行代码,静态分析pytorch模型的错误
  • 原文地址:https://www.cnblogs.com/fatttcat/p/10021709.html
Copyright © 2020-2023  润新知