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  • 207. Course Schedule

    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, is it possible for you to finish all courses?

    For example:

    2, [[1,0]]

    There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

    2, [[1,0],[0,1]]

    There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

    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.

    题目含义: 有从0到n-1共n门课程,且一门课程会有前导课程。例如2, [[1,0]],要学习课程2先要学习课程1,要学习课程1先要学习课程0.。判断schedule是否正确。

     1     public boolean canFinish(int numCourses, int[][] prerequisites) {
     2  //     要上prerequisites[i][0]的课,必须先上prerequisites[i][1]的课 题目是要判断给定的序列是否有环路,有的话返回false
     3         // 参数检查
     4         if (prerequisites == null) {
     5             return false;
     6         }
     7         int len = prerequisites.length;
     8         if (numCourses <= 0 || len == 0) {
     9             return true;
    10         }
    11 
    12         // 记录每个course的prerequisites的数量
    13         int[] pCounter = new int[numCourses];
    14         for (int i = 0; i < len; i++) {
    15             pCounter[prerequisites[i][0]]++;  //prerequisites[i][0]这些课程都有前缀课程prerequisites[i][1]
    16         }
    17         // 用队列记录可以直接访问的course
    18         LinkedList<Integer> queue = new LinkedList<Integer>(); //具备上课条件,可以上课的集合
    19         for (int i = 0; i < numCourses; i++) {
    20             if (pCounter[i] == 0) {
    21                 queue.add(i);
    22             }
    23         }
    24 
    25         // 取出队列的course,判断
    26         int numNoPre = queue.size();
    27         while (!queue.isEmpty()) {
    28             int top = queue.remove();
    29             for (int i = 0; i < len; i++) {
    30                 // 该course是某个course的prerequisites
    31                 if (prerequisites[i][1] == top) {
    32                     pCounter[prerequisites[i][0]]--; //那门课程的前缀课程数可以减一,因为top课程可以上
    33                     if (pCounter[prerequisites[i][0]] == 0) {
    34                         numNoPre++; //那门课程的前缀课程都可以上了,所以那门课程也可以上了
    35                         queue.add(prerequisites[i][0]);  //将那门课程也加到可以上课的队列中
    36                     }
    37                 }
    38             }
    39         }
    40         return numNoPre == numCourses;       
    41     }
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  • 原文地址:https://www.cnblogs.com/wzj4858/p/7723443.html
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