zoukankan      html  css  js  c++  java
  • 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:
    约瑟夫解决问题的循环链表
    [Erlang危机](5.1.0)VM检测概述
    找呀志_java网络编程(5)TCP和udp差额
    有序输出两棵二叉查找树中的元素
    1234567选择3个数字组合
    八皇后问题
    矩阵乘法运算
    求π的近似值
    证明不小于6的偶数都是两个素数之和
    最大公约数最小公倍数
  • 原文地址:https://www.cnblogs.com/fatttcat/p/10021709.html
Copyright © 2011-2022 走看看