LeetCode 210 Course Schedule II

题目描述

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:

[code]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]

[code]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].

Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

分析

参考：LeetCode 207 Course Schedule

只是加了一句话而已，用于保存结果。注意prerequisites为空的时候，任意输出一组结果即可。

代码

[code]    public static int[] findOrder(int numCourses, int[][] prerequisites) {

        // 参数检查
        if (prerequisites == null) {
            throw new IllegalArgumentException();
        }

        int len = prerequisites.length;
        if (len == 0) {
            int[] seq = new int[numCourses];
            for (int i = 0; i < seq.length; i++) {
                seq[i] = i;
            }
            return seq;
        }

        int[] seq = new int[numCourses];
        int c = 0;

        // 记录每个course的prerequisites的数量
        int[] pCounter = new int[numCourses];
        for (int i = 0; i < len; i++) {
            pCounter[prerequisites[i][0]]++;
        }

        // 用队列记录可以直接访问的course
        LinkedList<Integer> queue = new LinkedList<Integer>();
        for (int i = 0; i < numCourses; i++) {
            if (pCounter[i] == 0) {
                queue.add(i);
            }
        }

        // 取出队列的course，判断
        int numNoPre = queue.size();
        while (!queue.isEmpty()) {
            int top = queue.remove();
            // 保存结果 +_+
            seq[c++] = top;
            for (int i = 0; i < len; i++) {
                // 该course是某个course的prerequisites
                if (prerequisites[i][1] == top) {
                    pCounter[prerequisites[i][0]]--;
                    if (pCounter[prerequisites[i][0]] == 0) {
                        numNoPre++;
                        queue.add(prerequisites[i][0]);
                    }
                }
            }
        }

        if (numNoPre != numCourses) {
            return new int[] {};
        }

        return seq;
    }