【LeetCode】Course Schedule II 解题报告

标签：topological sort   graph   bfs   leetcode   

【题目】

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.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

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].

相比较于 【LeetCode】Course Schedule 解题报告，返回的结果是一组排序结果，而非是否能够排序。

因为只要返回任意一种合法的拓扑排序结果即可，所以只需在BFS过程中记录下来先后访问的节点即可。

【Java代码】

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        List<Set<Integer>> adjLists = new ArrayList<Set<Integer>>();
        for (int i = 0; i < numCourses; i++) {
            adjLists.add(new HashSet<Integer>());
        }
        
        for (int i = 0; i < prerequisites.length; i++) {
            adjLists.get(prerequisites[i][1]).add(prerequisites[i][0]);
        }
        
        int[] indegrees = new int[numCourses];
        for (int i = 0; i < numCourses; i++) {
            for (int x : adjLists.get(i)) {
                indegrees[x]++;
            }
        }
        
        Queue<Integer> queue = new LinkedList<Integer>();
        for (int i = 0; i < numCourses; i++) {
            if (indegrees[i] == 0) {
                queue.offer(i);
            }
        }
        
        int[] res = new int[numCourses];
        int count = 0;
        while (!queue.isEmpty()) {
            int cur = queue.poll();
            for (int x : adjLists.get(cur)) {
                indegrees[x]--;
                if (indegrees[x] == 0) {
                    queue.offer(x);
                }
            }
            res[count++] = cur;
        }
        
        if (count == numCourses) return res;
        return new int[0];
    }
}

【LeetCode】Course Schedule II 解题报告

标签：topological sort   graph   bfs   leetcode   

原文地址：http://blog.csdn.net/ljiabin/article/details/45847019