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

Hints:

• This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
• Topological Sort via DFS – A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
• Topological sort could also be done via BFS.

分析

题目等价为：检测图中是否有环

参考网址：LeetCode – Course Schedule (Java)

代码

BFS：

    // BFS
    public static boolean canFinish(int numCourses, int[][] prerequisites) {

        // 参数检查
        if (prerequisites == null) {
            return false;
        }

        int len = prerequisites.length;
        if (numCourses <= 0 || len == 0) {
            return true;
        }

        // 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();
            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]);
                    }
                }
            }
        }

        return numNoPre == numCourses;
    }

DFS：

    // DFS
    public static boolean canFinish2(int numCourses, int[][] prerequisites) {

        // 参数检查
        if (prerequisites == null) {
            return false;
        }

        int len = prerequisites.length;
        if (numCourses <= 0 || len == 0) {
            return true;
        }

        int[] visit = new int[numCourses];

        // key：course；value：以该course为prerequisites的course
        HashMap<Integer, ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();

        // 初始化map
        for (int[] p : prerequisites) {
            if (map.containsKey(p[1])) {
                map.get(p[1]).add(p[0]);
            } else {
                ArrayList<Integer> l = new ArrayList<Integer>();
                l.add(p[0]);
                map.put(p[1], l);
            }
        }

        // dfs
        for (int i = 0; i < numCourses; i++) {
            if (!canFinishDFS(map, visit, i)) {
                return false;
            }
        }

        return true;
    }

    private static boolean canFinishDFS(
            HashMap<Integer, ArrayList<Integer>> map, int[] visit, int i) {

        if (visit[i] == -1) {
            return false;
        }

        if (visit[i] == 1) {
            return true;
        }

        visit[i] = -1;

        // course i是某些course的prerequisites
        if (map.containsKey(i)) {
            for (int j : map.get(i)) {
                if (!canFinishDFS(map, visit, j)) {
                    return false;
                }
            }
        }

        visit[i] = 1;

        return true;
    }