210. Course Schedule II

短命女 2022-04-24 15:38 188阅读 0赞

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

与上一题[207. Course Schedule][]类似，不过本题需要返回一个学习课程的顺序，对上一题的BFS方法稍微修改即可。代码见底部。

PS：210. Course Schedule II的[Solution][]写得挺有意思的。

public int[] findOrder(int numCourses, int[][] prerequisites) {
            boolean[][] isPre = new boolean[numCourses][numCourses];//isPre[i][j]为真表示课程i是课程j的前置课程
            int[] value = new int[numCourses];//value[i]表示课程i所需的前置课程数
            for(int i = 0; i < prerequisites.length; ++i) {
                int pre = prerequisites[i][1];
                int ready = prerequisites[i][0];
                if(!isPre[pre][ready]) {
                    value[ready]++;
                }
                isPre[pre][ready] = true;
            }
            int[] ans = new int[numCourses];
            int count = 0;
            Queue<Integer> queue = new ArrayDeque<>();
            for(int i = 0; i < value.length; ++i) {
                if(value[i] == 0) {
                    queue.add(i);
                }
            }
            while(!queue.isEmpty()) {
                int course = queue.poll();
                ans[count++] = course;
                for(int i = 0; i < numCourses; ++i) {
                    if(isPre[course][i] && --value[i] == 0) {
                        queue.add(i);
                    }
                }
            }
            return count != numCourses ? new int[0] : ans;
        }

[207. Course Schedule]: https://blog.csdn.net/weixin_41704428/article/details/89673065
[Solution]: https://leetcode.com/problems/course-schedule-ii/solution/