Course Schedule I && II 总结

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.

    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<List<Integer>> g = new ArrayList<>();
        for (int i = 0; i < numCourses; i++) {
            g.add(new ArrayList<>());
        }
        for (int i = 0; i < prerequisites.length; i++) {
            g.get(prerequisites[i][1]).add(prerequisites[i][0]);
        }
        boolean[] visited = new boolean[numCourses];
        for (int i = 0; i < numCourses; i++) {
            if (!dfs(visited, g, i)) return false;
        }
        return true;
    }
    
    private boolean dfs(boolean[] visited, List<List<Integer>> g, int n) {
        if (visited[n]) return false;
        else visited[n] = true;
        List<Integer> neighbors = g.get(n);
        for (int v : neighbors) {
            if (!dfs(visited, g, v)) return false;
        }
        visited[n] = false;
        return true;
    }

public boolean canFinish(int numCourses, int[][] prerequisites) {
    
        List<Integer>[] matrix = new List[numCourses];
        int[] indegree = new int[numCourses];

        // E part
        for (int[] pre : prerequisites) {
            int preCourse = pre[1];
            int readyCourse = pre[0];
            List<Integer> list = matrix[preCourse];
            if (list == null) {
                list = new LinkedList<>();   
                matrix[preCourse] = list;
            }
            list.add(readyCourse);
            indegree[readyCourse]++;
        }

        Queue<Integer> queue = new LinkedList<>();
        for (int i=0; i<numCourses; i++) {
            if (indegree[i] == 0) queue.offer(i);
        }
        int count = 0;
        // V part
        while (!queue.isEmpty()) {
            int vertex = queue.poll();
            count++;
            List<Integer> adjacent = matrix[vertex];
            if (adjacent == null) continue;
            for (int neighbor : adjacent) {
                indegree[neighbor]--;
                if (indegree[neighbor] == 0)
                    queue.offer(neighbor);
            }
        }
        return count == numCourses;
    }

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].

用 adjacent list 来表示 图， 计算出每个点的 indegree， 到这个点的 edge 的 数目

把 indegree 是 0 的点加入到 queue 中， 依次添加。

public int[] findOrder(int numCourses, int[][] prerequisites) {  
        List<List<Integer>> adjLists = new ArrayList<List<Integer>>();  
        for (int i = 0; i < numCourses; i++) {  
            adjLists.add(new ArrayList<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];  
    } 

1 -- visited

2 -- tested

public class Graph{
        LinkedList<Integer> adj[];
        public Graph(int v){
            adj = new LinkedList[v];
            for(int i=0; i<v; i++){
                adj[i] = new LinkedList<>();
            }
        }
        public void addEdge(int s, int d){
            adj[s].add(d);
        }
    }
    
    public boolean topSort(Graph g, int v, Stack<Integer> stack, int[] visited){
        if(visited[v] == 1)
            return false;
        if(visited[v] == 2)
            return true;
        visited[v] = 1;
        for(int i=0; i<g.adj[v].size(); i++){
            if(!topSort(g, g.adj[v].get(i), stack, visited)){
                return false;
            }
        }
        stack.add(v);
        visited[v] = 2;
        return true;
    }
    
    public int[] findOrder(int n, int[][] reqs) {
        if(n==0)
            return null;

        Graph g = new Graph(n);
        int i=0;
        if(reqs != null){
            for(int[] req: reqs)
                g.addEdge(req[1], req[0]);
        }
        Stack<Integer> stack = new Stack<>();
        int[] visited = new int[n];
        for(; i<n; i++){
            if(!topSort(g, i, stack, visited))
                return new int[0];
        }
        int[] result = new int[n];
        i = 0;
        while(!stack.isEmpty())
            result[i++] = stack.pop();
        return result;
    }